Answer key · Unit 10

Redox and electrochemistry.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 25 minutes

Oxidation and reduction

Reactions in which electrons pass from one substance to another: who loses and who gains, where the electrons go in a cell, oxidation numbers, and why oxidation never happens alone.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
Losing electrons
Practice · Choose
Question
In the cell the zinc strip is slowly eaten away: $\mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^-$ ⁠. What is happening to the zinc?

Correct answer
- Correct It is oxidised: it loses electrons
Setup, choices and feedback
The Scene at this step
Load resistance
50 Ω
Left metal
Zinc
Show ions and electrons
Off
Choices
Shown in a random order.
- Correct It is oxidised: it loses electrons
- Wrong It is reduced: it loses electrons
  Misconception: Swaps oxidation and reduction: takes gaining electrons to be oxidation.
  Feedback from the catalogue
  Oxidation is losing electrons, and reduction is gaining electrons. Zinc that becomes $\mathrm{Zn^{2+}}$ ⁠ has lost two electrons: it has been oxidised, and its charge has gone up. A copper ion that becomes copper has gained two electrons: it has been reduced, and its charge has gone down, hence the name. A memory aid: OIL RIG, oxidation is loss, reduction is gain.
- Wrong It is oxidised: it gains electrons
  Misconception: Swaps oxidation and reduction: takes gaining electrons to be oxidation.
  Feedback from the catalogue
  Oxidation is losing electrons, and reduction is gaining electrons. Zinc that becomes $\mathrm{Zn^{2+}}$ ⁠ has lost two electrons: it has been oxidised, and its charge has gone up. A copper ion that becomes copper has gained two electrons: it has been reduced, and its charge has gone down, hence the name. A memory aid: OIL RIG, oxidation is loss, reduction is gain.
- Wrong It only dissolves, like salt in water, with no chemical change
  Feedback
  Dissolving salt was ions already in the crystal. Here neutral zinc atoms become ions: they change their number of electrons, and that is a chemical reaction.
Feedback for a right answer
Each zinc atom loses two electrons and becomes a $\mathrm{Zn^{2+}}$ ⁠ ion that goes into the solution. Losing electrons is oxidation. The electrons do not vanish: they leave along the wire for the copper strip.

Worked solution
1. On the left of the arrow is a neutral atom, $\mathrm{Zn}$ ⁠; on the right a $\mathrm{Zn^{2+}}$ ⁠ ion and two electrons.
2. The zinc lost two electrons, and its charge rose from 0 to +2.
3. Losing electrons is oxidation.
Step 2 of 4
Open this step
Where the electrons go
Practice · Predict, then look
Question
Show ions and electrons will reveal the ions and the electrons. Which way do the electrons flow?

Correct answer
- Correct Along the wire, from the zinc to the copper
Setup, choices and feedback
The Scene at this step
Load resistance
50 Ω
Left metal
Zinc
Show ions and electrons
Offfixed
Choices
- Correct Along the wire, from the zinc to the copper
- Wrong Along the wire, from the copper to the zinc
  Misconception: Swaps oxidation and reduction: takes gaining electrons to be oxidation.
  Feedback from the catalogue
  Oxidation is losing electrons, and reduction is gaining electrons. Zinc that becomes $\mathrm{Zn^{2+}}$ ⁠ has lost two electrons: it has been oxidised, and its charge has gone up. A copper ion that becomes copper has gained two electrons: it has been reduced, and its charge has gone down, hence the name. A memory aid: OIL RIG, oxidation is loss, reduction is gain.
- Wrong Through the salt bridge, from beaker to beaker
  Misconception: Thinks electrons flow through the solutions and the salt bridge to close the circuit.
  Feedback from the catalogue
  Electrons flow only through metal: the wire and the electrodes. In the solutions and the salt bridge there are no free electrons; there the charge is carried by ions. Positive ions move towards the cathode and negative ions towards the anode, so each solution stays neutral and the circuit is complete.
Once the prediction is locked in, the Scene changes
Show ions and electrons
Onfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The zinc is oxidised and gives up electrons, which travel along the wire to the copper. There $\mathrm{Cu^{2+}}$ ⁠ ions from the solution take them and become copper. In the salt bridge it is ions that move, not electrons.

Worked solution
1. The zinc is oxidised: $\mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^-$ ⁠, so electrons are made at the zinc strip.
2. The copper ions are reduced: $\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}$ ⁠, so electrons are used up at the copper strip.
3. So the electrons travel from the zinc to the copper, through the only metal that joins them: the wire.
Step 3 of 4
Open this step
Oxidation numbers
Practice · Calculate
Question
What is the oxidation number of sulfur in the sulfate ion, $\mathrm{SO_4^{2-}}$ ⁠? Answer with a signed number.

Correct answer
6 ± 0.01
Setup, choices and feedback
The Scene at this step
Load resistance
50 Ω
Left metal
Zinc
Show ions and electrons
On
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- -2 ± 0.01
  Misconception: Gives an atom in a compound the charge of its usual simple ion, instead of working the oxidation number out.
  Feedback from the catalogue
  An oxidation number is worked out from the compound itself, not from the element’s usual ion. For sulfur in $\mathrm{SO_4^{2-}}$ ⁠: each oxygen is −2, four oxygens are −8, and the total must equal the ion’s charge, −2. So sulfur is +6, not −2 as in the sulfide ion $\mathrm{S^{2-}}$ ⁠.
Feedback for a right answer
$x + 4 \times (-2) = -2$ ⁠, so $x = +6$ ⁠. Sulfur has six outer electrons, and in sulfate it is as if it has handed all of them to the oxygens.

Hint
Oxygen is −2, and the oxidation numbers add up to the ion’s charge.

Worked solution
1. Oxygen’s oxidation number in compounds: −2.
2. The numbers in an ion add up to its charge: $x + 4 \times (-2) = -2$ ⁠.
3. $x - 8 = -2$ ⁠, so $x = +6$ ⁠.
If you answered −2, you gave sulfur the charge of the sulfide ion, $\mathrm{S^{2-}}$ ⁠. In sulfate it is bonded to oxygens, which pull electrons harder than it does.
Step 4 of 4
Open this step
Always together
Practice · Choose
Question
Iron rusts in damp air: $4\mathrm{Fe} + 3\mathrm{O_2} \rightarrow 2\mathrm{Fe_2O_3}$ ⁠. The iron is oxidised. What happens to the oxygen?

Correct answer
- Correct It is reduced: each oxygen atom gains two electrons
Setup, choices and feedback
The Scene at this step
Load resistance
50 Ω
Left metal
Zinc
Show ions and electrons
On
Choices
Shown in a random order.
- Correct It is reduced: each oxygen atom gains two electrons
- Wrong Nothing: oxidation is a reaction of the iron alone
  Misconception: Thinks an oxidation can happen on its own, without a reduction.
  Feedback from the catalogue
  Electrons do not vanish. Every electron one substance loses, another must gain. So oxidation and reduction always happen together, and the electrons lost equal the electrons gained. That is why it is called redox, reduction and oxidation as one.
- Wrong It is oxidised too
  Misconception: Swaps oxidation and reduction: takes gaining electrons to be oxidation.
  Feedback from the catalogue
  Oxidation is losing electrons, and reduction is gaining electrons. Zinc that becomes $\mathrm{Zn^{2+}}$ ⁠ has lost two electrons: it has been oxidised, and its charge has gone up. A copper ion that becomes copper has gained two electrons: it has been reduced, and its charge has gone down, hence the name. A memory aid: OIL RIG, oxidation is loss, reduction is gain.
Feedback for a right answer
Oxygen’s oxidation number falls from 0 in $\mathrm{O_2}$ ⁠ to −2 in the oxide: it has been reduced. The iron rises from 0 to +3. The iron is the reducing agent, the substance that gives electrons, and the oxygen is the oxidising agent, which takes them.

Worked solution
1. In $\mathrm{Fe}$ ⁠ and $\mathrm{O_2}$ ⁠ the oxidation numbers are 0: elements.
2. In $\mathrm{Fe_2O_3}$ ⁠: oxygen −2, so $2x + 3(-2) = 0$ ⁠, $x = +3$ ⁠.
3. The iron lost electrons (0 → +3) and the oxygen gained them (0 → −2).
4. 4 iron atoms give 12 electrons, and 6 oxygen atoms take 12.

Lesson 2 of 4 · About 25 minutes

The reactivity series

Why sodium skates over water and potassium catches fire, how metals are ranked by how readily they lose electrons, and what the ranking predicts: which metal pushes which out of solution, and which gives off hydrogen in acid.

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The 6 steps and their answers

Step 1 of 6
Open this step
Down the group
Practice · Predict, then look
Question
In the left trough sodium skates over the water, fizzing. What will potassium, below sodium in the table, do?

Correct answer
- Correct React more violently: its hydrogen will catch fire
Setup, choices and feedback
The Scene at this step
Row of the table
Na · Mgfixed
Water temperature
293.1 K
Zoom to the atoms
Troughs
Choices
- Correct React more violently: its hydrogen will catch fire
- Wrong React more gently
  Misconception: Thinks group 1 and 2 metals get less reactive down the group, as the halogens do.
  Feedback from the catalogue
  For the metals reactivity rises down the group. A metal reacts by losing electrons, and the bigger the atom, the further the outer electron is from the nucleus and the more loosely it is held. Lithium fizzes slowly, sodium skates on the water, potassium catches fire, and caesium explodes. For the halogens, which react by gaining an electron, the trend runs the other way.
- Wrong React just like sodium, since both have one outer electron
  Feedback
  Both have one outer electron, so they react in the same way. But how easily it is lost depends on its distance from the nucleus.
Once the prediction is locked in, the Scene changes
Row of the table
K · Cafixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Potassium reacts so fast that the heat lights the hydrogen in a lilac flame. $2\mathrm{K} + 2\mathrm{H_2O} \rightarrow 2\mathrm{KOH} + \mathrm{H_2}$ ⁠: the same reaction as sodium’s, only faster.

Worked solution
1. Metals react by losing their outer electrons.
2. Potassium has one more shell than sodium, so its outer electron is further from the nucleus.
3. It is held more loosely and lost more easily: potassium is more reactive.
Step 2 of 6
Open this step
Why
Practice · Choose
Question
Why does caesium react with water far more violently than lithium?

Correct answer
- Correct Its outer electron is far from the nucleus and held loosely
Setup, choices and feedback
The Scene at this step
Row of the table
Cs · Ba
Water temperature
293.1 K
Zoom to the atoms
Atoms
Choices
Shown in a random order.
- Correct Its outer electron is far from the nucleus and held loosely
- Wrong It has more outer electrons to give away
  Feedback
  Lithium and caesium both have exactly one outer electron: both are in group 1.
- Wrong It is heavier, so it hits the water harder
  Feedback
  Mass does not set how fast it reacts. What does is how easily the atom gives up its electron.
- Wrong Its nucleus is stronger and pushes the electron out
  Feedback
  A positive nucleus pulls electrons, it does not push them. In caesium its pull reaches the outer electron weakly, through five inner shells.
Feedback for a right answer
Caesium has six shells, and its outer electron feels a weak pull from the nucleus, screened by the inner electrons. So it takes only 376 kJ/mol to remove it, against 520 kJ/mol for lithium.

Worked solution
1. All group 1 metals react by losing their single outer electron.
2. Going down the group, shells are added, and the outer electron is further from the nucleus.
3. It is pulled more weakly, so it is lost more easily, and the reaction is faster.
Step 3 of 6
Open this step
Group 2
Practice · Predict, then look
Question
In the right trough magnesium sits in cold water, and almost nothing happens. What will happen when you heat the water to 80 °C?

Correct answer
- Correct It will start to fizz slowly
Setup, choices and feedback
The Scene at this step
Row of the table
Na · Mgfixed
Water temperature
293.1 Kfixed
Zoom to the atoms
Troughs
Choices
- Correct It will start to fizz slowly
- Wrong Nothing: magnesium does not react with water
  Feedback
  Magnesium does react with water, but very slowly in the cold. Heat speeds every reaction up.
- Wrong It will react as violently as sodium
  Feedback
  Magnesium has to lose two electrons, held more tightly than sodium’s one. Heat speeds it up, but not that much.
Once the prediction is locked in, the Scene changes
Water temperature
353.1 Kfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
In hot water magnesium fizzes slowly, giving off hydrogen, and the indicator turns pink round it: $\mathrm{Mg} + 2\mathrm{H_2O} \rightarrow \mathrm{Mg(OH)_2} + \mathrm{H_2}$ ⁠. Group 2 metals are less reactive than their group 1 neighbours in the same row.

Worked solution
1. Magnesium is a reactive metal, but less so than sodium: it has to give up two electrons.
2. In cold water the reaction is too slow to see.
3. Heating speeds the reaction up, and bubbles of hydrogen appear.
Step 4 of 6
Open this step
Displacement
Practice · Choose
Question
A zinc strip is dipped in blue copper sulfate solution, $\mathrm{CuSO_4}$ ⁠. What will you see after a few minutes?

Correct answer
- Correct A reddish-brown coat of copper on the zinc, and the blue fading
Setup, choices and feedback
The Scene at this step
Row of the table
K · Ca
Water temperature
293.1 K
Zoom to the atoms
Troughs
Choices
Shown in a random order.
- Correct A reddish-brown coat of copper on the zinc, and the blue fading
- Wrong Nothing: a metal does not react with a solution of another metal
  Feedback
  Zinc is more reactive than copper, so it gives electrons to the copper ions.
- Wrong The zinc coats itself with more zinc from the solution
  Feedback
  At first the solution has no zinc ions; it has copper ions. They are what gets reduced.
Feedback for a right answer
$\mathrm{Zn} + \mathrm{Cu^{2+}} \rightarrow \mathrm{Zn^{2+}} + \mathrm{Cu}$ ⁠: the zinc is oxidised and goes into solution, and the copper ions are reduced and coat it. This is a displacement reaction: a more reactive metal pushes a less reactive one out of solution. It is the galvanic cell, without the wire.

Worked solution
1. In the reactivity series zinc is above copper: it gives up electrons more readily.
2. So the zinc is oxidised, $\mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^-$ ⁠, and the copper ions are reduced, $\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}$ ⁠.
3. Copper settles on the zinc, and the blue copper ions in the solution thin out.
Step 5 of 6
Open this step
Not the other way
Practice · Choose
Question
And what will happen to a silver wire in the same copper sulfate solution?

Correct answer
- Correct Nothing
Setup, choices and feedback
The Scene at this step
Row of the table
Na · Mgdefault
Water temperature
293.1 Kdefault
Zoom to the atoms
Troughsdefault
Choices
Shown in a random order.
- Correct Nothing
- Wrong It will be coated with copper
  Misconception: Thinks any metal will push any other metal out of a solution of its salt.
  Feedback from the catalogue
  A metal pushes another metal out of solution only if it is more reactive, more ready to lose electrons. Zinc, above copper in the reactivity series, gets coated with copper in copper sulfate solution. Silver, below copper, does nothing in it.
- Wrong It will dissolve, and the blue will deepen
  Misconception: Thinks any metal will push any other metal out of a solution of its salt.
  Feedback from the catalogue
  A metal pushes another metal out of solution only if it is more reactive, more ready to lose electrons. Zinc, above copper in the reactivity series, gets coated with copper in copper sulfate solution. Silver, below copper, does nothing in it.
Feedback for a right answer
Silver is below copper in the reactivity series: it holds its electrons more tightly than copper does, so it will not give them to the copper ions. The other way round does work: a copper wire in silver nitrate solution grows silver crystals.

Worked solution
1. To get coated with copper, the silver would have to give electrons to the copper ions.
2. Silver is below copper in the reactivity series: it is less ready to lose electrons.
3. So there is no reaction.
Step 6 of 6
Open this step
Metals and acids
Practice · Choose
Question
Which of these metals will give off hydrogen in dilute acid? Choose all of them.

Correct answer
- Correct Magnesium
- Correct Zinc
- Correct Iron
Setup, choices and feedback
The Scene at this step
Row of the table
Na · Mgdefault
Water temperature
293.1 Kdefault
Zoom to the atoms
Troughsdefault
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct Magnesium
- Correct Zinc
- Correct Iron
- Wrong Copper
  Misconception: Thinks any metal will push any other metal out of a solution of its salt.
  Feedback from the catalogue
  A metal pushes another metal out of solution only if it is more reactive, more ready to lose electrons. Zinc, above copper in the reactivity series, gets coated with copper in copper sulfate solution. Silver, below copper, does nothing in it.
- Wrong Silver
  Misconception: Thinks any metal will push any other metal out of a solution of its salt.
  Feedback from the catalogue
  A metal pushes another metal out of solution only if it is more reactive, more ready to lose electrons. Zinc, above copper in the reactivity series, gets coated with copper in copper sulfate solution. Silver, below copper, does nothing in it.
Feedback for a right answer
Metals above hydrogen in the reactivity series give electrons to the acid’s $\mathrm{H^+}$ ⁠ ions: $\mathrm{Zn} + 2\mathrm{H^+} \rightarrow \mathrm{Zn^{2+}} + \mathrm{H_2}$ ⁠. Copper and silver are below hydrogen, and do not react. That is why coins and jewellery are made of them.

Worked solution
1. The reactivity series: $\mathrm{K, Na, Ca, Mg, Al, Zn, Fe, Pb, (H), Cu, Ag, Au}$ ⁠.
2. A metal releases hydrogen from an acid only if it is above hydrogen.
3. Magnesium, zinc and iron are above it; copper and silver below.

Lesson 3 of 4 · About 30 minutes

Galvanic cells

How a redox reaction makes electricity: anode and cathode, voltage from standard potentials, why the poles can swap, how much voltage is lost inside the cell, and what the salt bridge does.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
A cell’s voltage
Practice · Calculate
Question
The standard reduction potentials are $E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34\ \mathrm{V}$ ⁠ and $E^\circ(\mathrm{Zn^{2+}/Zn}) = -0.76\ \mathrm{V}$ ⁠. What is the voltage of a zinc–copper cell with no current flowing?

Correct answer
1.1 V ± 0.01 V
Setup, choices and feedback
The Scene at this step
Load resistance
500 Ω
Left metal
Zincfixed
Show ions and electrons
Off
Accepted within
± 0.01 V
Computed from the Scene at this step’s setting
Cell voltage with no current: positive when the left metal is the − terminal/nodes/cell/outputs/emf
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- -0.42 V ± 0.01 V
  Misconception: Adds the two standard electrode potentials instead of subtracting them.
  Feedback from the catalogue
  A cell’s voltage is the difference of the potentials: $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ ⁠. For a zinc–copper cell: $0.34 - (-0.76) = 1.10\ \mathrm{V}$ ⁠. Adding gives $0.34 + (-0.76) = -0.42$ ⁠, and a negative voltage for a working cell is a sign something has flipped.
Feedback for a right answer
$0.34 - (-0.76) = 1.10\ \mathrm{V}$ ⁠. The copper is the cathode, the positive terminal, and the zinc is the anode, the negative one. That is the voltage of the Daniell cell, one of the first batteries.

Hint
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ ⁠.

Worked solution
1. The cathode is the metal with the higher potential: copper, +0.34 V. The anode: zinc, −0.76 V.
2. $E^\circ_{\text{cell}} = 0.34 - (-0.76)$ ⁠.
3. $= 0.34 + 0.76 = 1.10$ ⁠.
4. The answer: 1.10 V.
If you got −0.42 V, you added the potentials instead of subtracting.
Step 2 of 5
Open this step
Choosing the metals
Practice · Reach the target
Question
Choose, with Left metal, the metal that gives the cell the highest voltage, then press Check.

Goal
- 2.71 V ± 0.01 V
  At the start of the step: 1.1 V
Setup, choices and feedback
The Scene at this step
Load resistance
500 Ωfixed
Left metal
Zinc
Show ions and electrons
Off
Read from the Scene at
Cell voltage with no current: positive when the left metal is the − terminal/nodes/cell/outputs/emf
One setting that reaches it
Left metal
Magnesium
Feedback for a right answer
Magnesium: $0.34 - (-2.37) = 2.71\ \mathrm{V}$ ⁠. The further apart the two metals are in the reactivity series, the higher the voltage.

Feedback for a wrong answer
Not yet. Find the metal with the lowest reduction potential.

Hint
The higher the left metal in the reactivity series, the harder it “pushes” electrons.

Worked solution
1. The copper on the right is fixed, +0.34 V.
2. $E^\circ_{\text{cell}} = 0.34 - E^\circ_{\text{left}}$ ⁠: largest when $E^\circ_{\text{left}}$ ⁠ is lowest.
3. Magnesium, −2.37 V, is the lowest of the five.
4. Choose magnesium: 2.71 V.
Step 3 of 5
Open this step
The poles swap
Practice · Predict, then look
Question
In the zinc–copper cell the zinc dissolves. Which strip will dissolve when you change the left metal to Silver?

Correct answer
- Correct The copper
Setup, choices and feedback
The Scene at this step
Load resistance
50 Ω
Left metal
Zincfixed
Show ions and electrons
On
Choices
- Correct The copper
- Wrong The silver
  Misconception: Thinks the electrode with the more positive potential is the one that dissolves.
  Feedback from the catalogue
  The electrode that dissolves is the one being oxidised: the metal that holds its electrons less tightly, with the lower potential. In a silver–copper cell, $E^\circ(\mathrm{Cu}) = +0.34$ ⁠ and $E^\circ(\mathrm{Ag}) = +0.80$ ⁠, so the copper dissolves and silver builds up.
- Wrong Neither: silver does not react, so no current flows
  Feedback
  Silver is not oxidised here, but silver ions can be reduced. So the copper is the one that gives up electrons.
Once the prediction is locked in, the Scene changes
Left metal
Silverfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Silver is below copper in the reactivity series, so now copper is the anode: it is oxidised, and silver ions are reduced onto the silver strip. The poles have swapped, and the voltage is $0.34 - 0.80 = -0.46\ \mathrm{V}$ ⁠: 0.46 V the other way round.

Worked solution
1. $E^\circ(\mathrm{Ag^+/Ag}) = +0.80$ ⁠, higher than $E^\circ(\mathrm{Cu^{2+}/Cu}) = +0.34$ ⁠.
2. The metal with the lower potential is oxidised: copper.
3. So the copper dissolves, $\mathrm{Cu} \rightarrow \mathrm{Cu^{2+}} + 2e^-$ ⁠, and silver builds up on the left strip.
Step 4 of 5
Open this step
Voltage under load
Practice · Calculate
Question
The zinc–copper cell, 1.10 V, has an internal resistance of 10 Ω. Load resistance is 20 Ω. What does the meter across the load read?

Correct answer
0.7333 V ± 0.01 V
Setup, choices and feedback
The Scene at this step
Load resistance
20 Ωfixed
Left metal
Zincfixed
Show ions and electrons
Off
Accepted within
± 0.01 V
Computed from the Scene at this step’s setting
What the meter reads across the load/nodes/cell/outputs/voltage
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.1 V ± 0.01 V
  Misconception: Thinks a cell delivers its full emf at its terminals whatever current it supplies.
  Feedback from the catalogue
  A cell has internal resistance: the solutions and the salt bridge. When current flows, part of the voltage is used up inside the cell, $Ir$ ⁠, and the meter shows only the rest: $V = \varepsilon - Ir$ ⁠. The heavier the load and the bigger the current, the lower the voltage at the terminals.
Feedback for a right answer
$I = \dfrac{1.10}{20 + 10} = 0.0367\ \mathrm{A}$ ⁠, and $V = 0.0367 \times 20 = 0.73\ \mathrm{V}$ ⁠. A third of the voltage went inside the cell, in the solutions and the salt bridge.

Hint
First the current: $I = \dfrac{\varepsilon}{R + r}$ ⁠. Then $V = IR$ ⁠.

Worked solution
1. The givens: $\varepsilon = 1.10\ \mathrm{V}$ ⁠, $r = 10\ \Omega$ ⁠, $R = 20\ \Omega$ ⁠.
2. The current: $I = \dfrac{\varepsilon}{R + r} = \dfrac{1.10}{30} = 0.0367\ \mathrm{A}$ ⁠.
3. The voltage across the load: $V = IR = 0.0367 \times 20 = 0.733$ ⁠.
4. The answer: 0.73 V.
If you answered 1.10 V, you left out the internal resistance: that is the voltage only when no current flows.
Step 5 of 5
Open this step
The salt bridge
Practice · Choose
Question
What does the salt bridge between the two beakers do?

Correct answer
- Correct It lets ions move, so each solution stays neutral
Setup, choices and feedback
The Scene at this step
Load resistance
50 Ω
Left metal
Zinc
Show ions and electrons
On
Choices
Shown in a random order.
- Correct It lets ions move, so each solution stays neutral
- Wrong It carries the electrons back from the copper to the zinc
  Misconception: Thinks electrons flow through the solutions and the salt bridge to close the circuit.
  Feedback from the catalogue
  Electrons flow only through metal: the wire and the electrodes. In the solutions and the salt bridge there are no free electrons; there the charge is carried by ions. Positive ions move towards the cathode and negative ions towards the anode, so each solution stays neutral and the circuit is complete.
- Wrong It supplies the cell’s energy
  Feedback
  The energy comes from the reaction between the zinc and the copper ions. Without the bridge the reaction stops, but the bridge itself does not react.
Feedback for a right answer
Without the bridge, $\mathrm{Zn^{2+}}$ ⁠ ions would pile up in the left beaker and a positive charge would build there, while the right one turned negative. Within a moment those charges would stop the current. Through the bridge negative ions move left and positive ions right, and the circuit stays closed.

Worked solution
1. In the zinc beaker positive ions are made; in the copper beaker positive ions are used up.
2. With no way between the beakers, the first would charge positive and the second negative, and the current would stop.
3. The salt bridge passes ions and balances the charges. Electrons go only through the wire.

Lesson 4 of 4 · About 40 minutes

Batteries and electrolysis

A battery is a galvanic cell you can run backwards: what runs down as it discharges, what charging does, and how electrolysis makes metals and plates objects with them, gram by gram.

Open the lesson

The 10 steps and their answers

Step 1 of 10
Open this step
A battery is a cell
Practice · Choose
Question
When a phone runs on its battery all day, what does the battery lose?

Correct answer
- Correct Chemical energy, which turns into electrical energy
Setup, choices and feedback
The Scene at this step
Current
1 C
Charge level
60 %
Zoom to the ions
Ions
Choices
Shown in a random order.
- Correct Chemical energy, which turns into electrical energy
- Wrong Electric charge: it runs out of electrons
  Misconception: Thinks a battery stores electric charge, so charging fills it with electrons from the charger.
  Feedback from the catalogue
  A battery stores chemical energy, not charge. Every electron that leaves one terminal comes back at the other, and the battery stays neutral all the way. Charging pushes the electrons and the lithium ions back to where their chemical energy is high: it runs the reaction backwards.
- Wrong Lithium, which is burned up as it is used
  Feedback
  The lithium only moves from side to side inside the battery. That is why it can be recharged hundreds of times.
Feedback for a right answer
The battery stays neutral the whole time: every electron that leaves the negative terminal comes back at the positive one. What runs down is the chemical energy: the lithium has moved from a high-energy place, in the graphite, to a low-energy place, in the oxide.

Worked solution
1. In a closed circuit the current is the same everywhere: whatever goes out comes back.
2. So the battery loses neither electrons nor charge.
3. It loses chemical energy, which the reaction turns into electrical work in the circuit.
Step 2 of 10
Open this step
Charging
Practice · Predict, then look
Question
The battery is discharging: lithium ions cross inside it from the graphite to the oxide. What will the ions do when you plug in a charger?

Correct answer
- Correct They will go back from the oxide to the graphite
Setup, choices and feedback
The Scene at this step
Current
1 Cfixed
Charge level
60 %
Zoom to the ions
Ions
Choices
- Correct They will go back from the oxide to the graphite
- Wrong The charger will add new ions and electrons to the battery
  Misconception: Thinks a battery stores electric charge, so charging fills it with electrons from the charger.
  Feedback from the catalogue
  A battery stores chemical energy, not charge. Every electron that leaves one terminal comes back at the other, and the battery stays neutral all the way. Charging pushes the electrons and the lithium ions back to where their chemical energy is high: it runs the reaction backwards.
- Wrong They will stay put, and the charger will just fill the battery with electricity
  Misconception: Thinks a battery stores electric charge, so charging fills it with electrons from the charger.
  Feedback from the catalogue
  A battery stores chemical energy, not charge. Every electron that leaves one terminal comes back at the other, and the battery stays neutral all the way. Charging pushes the electrons and the lithium ions back to where their chemical energy is high: it runs the reaction backwards.
Once the prediction is locked in, the Scene changes
Current
-1 Cfixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The charger pushes electrons into the graphite, and the lithium ions come back to it through the electrolyte: the reaction runs backwards. No ion has been added and none lost; what has been added is chemical energy.

Worked solution
1. In discharge the chemical reaction runs by itself and drives electrons round the circuit: a galvanic cell.
2. In charging an outside supply drives the electrons the other way, forcing the reaction to run backwards.
3. So the lithium ions go back to the graphite, to the state of high chemical energy.
Step 3 of 10
Open this step
Electrolysis
Practice · Choose
Question
In the electrolysis of molten sodium chloride a current is passed between two electrodes. What forms at the negative electrode?

Correct answer
- Correct Sodium metal
Setup, choices and feedback
The Scene at this step
Current
-1 C
Charge level
60 %
Zoom to the ions
Ions
Choices
Shown in a random order.
- Correct Sodium metal
- Wrong Chlorine gas
  Misconception: Thinks the cathode is always the positive electrode.
  Feedback from the catalogue
  The cathode is the electrode where reduction happens, and the anode is where oxidation happens, always. The sign depends on the kind of cell: in a galvanic cell the cathode is the positive terminal, but in electrolysis the supply pushes electrons onto the cathode, and it is the negative one. There positive ions, such as $\mathrm{Na^+}$ ⁠, gain electrons.
- Wrong Hydrogen gas
  Feedback
  Molten salt has no water in it, so there is nowhere for hydrogen to come from. Hydrogen appears in the electrolysis of a solution in water.
Feedback for a right answer
The supply pushes electrons onto the negative electrode, which is the cathode: $\mathrm{Na^+}$ ⁠ ions are drawn to it and reduced, $\mathrm{Na^+} + e^- \rightarrow \mathrm{Na}$ ⁠. At the anode, the positive one, $\mathrm{Cl^-}$ ⁠ ions are oxidised to chlorine gas. That is how sodium is made industrially.

Worked solution
1. The negative electrode attracts positive ions: $\mathrm{Na^+}$ ⁠.
2. It is full of electrons from the supply, and the ions take them: reduction.
3. $\mathrm{Na^+} + e^- \rightarrow \mathrm{Na}$ ⁠: sodium metal forms.
If you chose chlorine, you assumed the cathode is always positive. In electrolysis the cathode is the negative terminal.
Step 4 of 10
Open this step
How much metal
Practice · Calculate
Question
A key is copper-plated: a current of 2.0 A passes through a $\mathrm{Cu^{2+}}$ ⁠ solution for 965 s. How many grams of copper build up on the key? Faraday’s constant $F = 96\,500\ \mathrm{C/mol}$ ⁠, and copper’s molar mass is 63.5 g/mol.

Correct answer
0.635 g ± 0.005 g
Setup, choices and feedback
The Scene at this step
Current
1 Cdefault
Charge level
60 %default
Zoom to the ions
Celldefault
Accepted within
± 0.005 g
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.27 g ± 0.005 g
  Misconception: Counts one electron per ion discharged, whatever the ion’s charge.
  Feedback from the catalogue
  Each ion takes as many electrons as its charge. $\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}$ ⁠: two moles of electrons for every mole of copper. So the same electric charge deposits half as many moles of copper as of silver, $\mathrm{Ag^+} + e^- \rightarrow \mathrm{Ag}$ ⁠.
Feedback for a right answer
$Q = 2.0 \times 965 = 1930\ \mathrm{C}$ ⁠, that is $0.020$ ⁠ mol of electrons, giving $0.010$ ⁠ mol of copper: 0.635 g.

Hint
$Q = It$ ⁠, moles of electrons $= \dfrac{Q}{F}$ ⁠, and two electrons for each copper ion.

Worked solution
1. The charge: $Q = It = 2.0 \times 965 = 1930\ \mathrm{C}$ ⁠.
2. Moles of electrons: $\dfrac{1930}{96\,500} = 0.020\ \mathrm{mol}$ ⁠.
3. $\mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu}$ ⁠: moles of copper $= \dfrac{0.020}{2} = 0.010$ ⁠.
4. The mass: $0.010 \times 63.5 = 0.635$ ⁠.
5. The answer: 0.635 g.
If you got 1.27 g, you counted one electron for each copper ion.
Step 5 of 10
Open this step
Counting charge
Practice · Calculate
Question
A phone battery supplies a current of 0.5 A for two hours. How much charge has passed round the circuit?

Correct answer
3600 C ± 1%
Setup, choices and feedback
The Scene at this step
Current
0.5 C
Charge level
60 %
Zoom to the ions
Cell
Accepted within
± 1%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1 C ± 0.01 C
  Misconception: Thinks a battery stores electric charge, so charging fills it with electrons from the charger.
  Feedback
  $0.5 \times 2 = 1$ ⁠ is ampere-hours, not coulombs. Convert hours to seconds. And note: this charge passed through the battery; it was not stored in it.
Feedback for a right answer
$Q = 0.5 \times 7200 = 3600\ \mathrm{C}$ ⁠: one ampere-hour. That charge went round the circuit and came back whole; what the battery gave was energy, about $3600 \times 3.7 \approx 13\,000\ \mathrm{J}$ ⁠ at 3.7 V.

Hint
$Q = It$ ⁠, with $t$ ⁠ in seconds.

Worked solution
1. $t = 2\ \mathrm{h} = 2 \times 3600 = 7200\ \mathrm{s}$ ⁠.
2. $Q = It = 0.5 \times 7200 = 3600$ ⁠.
3. The answer: 3600 C.
Step 6 of 10
Open this step
A faster current
Practice · Calculate
Question
The battery’s capacity is 3 Ah. A Current of 1C means a current that empties it in one hour. The current is now locked at 2C. How many amperes flow?

Correct answer
6 A ± 2%
Setup, choices and feedback
The Scene at this step
Current
2 Cfixed
Charge level
60 %
Zoom to the ions
Cell
Accepted within
± 2%
Computed from the Scene at this step’s setting
Current out of the + terminal; negative while charging/nodes/cell/outputs/batteryCurrent
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.5 A ± 2%
  Misconception: Divides the capacity by the C-rate to get the current, so a higher C-rate seems to mean a smaller current.
  Feedback from the catalogue
  The C-rate says how fast the battery empties: 1C in an hour, 2C in half an hour. Emptying it faster takes a bigger current, so the current is the capacity times the C-rate, per hour.
Feedback for a right answer
6 A: the battery will be empty in half an hour.

Hint
At 1C the battery empties in an hour: the current is the capacity divided by one hour.

Worked solution
1. At 1C: $\dfrac{3\ \mathrm{Ah}}{1\ \mathrm{h}} = 3$ ⁠ amperes.
2. At 2C: twice that, $6$ ⁠ amperes.
3. At that current the battery empties in half an hour.
If you answered 1.5 A, you halved instead of doubling. 2C empties it twice as fast, so the current is twice as big.
Step 7 of 10
Open this step
At the terminals
Practice · Calculate
Question
With no current flowing, the battery at 60% reads 3.80 V. Now it supplies 3 A, and its internal resistance is 0.05 Ω. What voltage reaches its terminals?

Correct answer
3.65 V ± 0.5%
Setup, choices and feedback
The Scene at this step
Current
1 Cfixed
Charge level
60 %fixed
Zoom to the ions
Cell
Accepted within
± 0.5%
Computed from the Scene at this step’s setting
Terminal voltage, the open-circuit voltage minus I·R/nodes/cell/outputs/batteryVoltage
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3.8 V ± 0.3%
  Misconception: Thinks a cell delivers its full emf at its terminals whatever current it supplies.
  Feedback from the catalogue
  A cell has internal resistance: the solutions and the salt bridge. When current flows, part of the voltage is used up inside the cell, $Ir$ ⁠, and the meter shows only the rest: $V = \varepsilon - Ir$ ⁠. The heavier the load and the bigger the current, the lower the voltage at the terminals.
Feedback for a right answer
3.65 V: 0.15 V goes on the resistance inside the battery.

Hint
Part of the voltage is dropped across the resistance inside the battery itself: $IR$ ⁠.

Worked solution
1. The drop inside: $IR = 3 \times 0.05 = 0.15$ ⁠ volts.
2. The voltage at the terminals: $3.80 - 0.15 = 3.65$ ⁠ volts.
If you answered 3.80 V, you forgot that the current flows through the inside of the battery too. The bigger the current, the lower the voltage at the terminals.
Step 8 of 10
Open this step
The power out
Practice · Calculate
Question
The Current is locked at 2C, 6 A. The battery reads 3.80 V with no current, and its internal resistance is 0.05 Ω. What power does it deliver to the bulb?

Correct answer
21 W ± 2%
Setup, choices and feedback
The Scene at this step
Current
2 Cfixed
Charge level
60 %fixed
Zoom to the ions
Cell
Accepted within
± 2%
Computed from the Scene at this step’s setting
Power delivered to the bulb; negative, taken from the charger/nodes/cell/outputs/batteryPower
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 22.8 W ± 1%
  Misconception: Thinks a cell delivers its full emf at its terminals whatever current it supplies.
  Feedback from the catalogue
  A cell has internal resistance: the solutions and the salt bridge. When current flows, part of the voltage is used up inside the cell, $Ir$ ⁠, and the meter shows only the rest: $V = \varepsilon - Ir$ ⁠. The heavier the load and the bigger the current, the lower the voltage at the terminals.
Feedback for a right answer
21 W. Another 1.8 W heats the battery from inside.

Hint
First find the voltage at the terminals, then $P = VI$ ⁠.

Worked solution
1. The drop inside: $6 \times 0.05 = 0.30$ ⁠ volts, so at the terminals $3.80 - 0.30 = 3.50$ ⁠ volts.
2. $P = VI = 3.50 \times 6 = 21$ ⁠ watts.
If you answered 22.8 W, you multiplied by the voltage with no current. The difference, 1.8 W, is the heat left inside the battery.
Step 9 of 10
Open this step
Heat inside
Practice · Predict, then look
Question
At a Current of 1C, 0.45 W of heat stays inside the battery. Predict: how much heat will stay in it when the current doubles to 2C?

Correct answer
- Correct Four times as much, 1.8 W
Setup, choices and feedback
The Scene at this step
Current
1 C
Charge level
60 %
Zoom to the ions
Ions
Choices
- Wrong The same, 0.45 W, because the resistance has not changed
  No feedback of its own; the learner sees the standard message
  Not yet. Look again and try once more.
- Wrong Twice as much, 0.9 W
  Misconception: Thinks the heat made in a resistance grows in step with the current.
  Feedback from the catalogue
  The heat in a resistor is $I^2 R$ ⁠. Twice the current means twice the charge each second, and each charge also loses twice the energy, because the drop $IR$ ⁠ doubles. Together: four times.
- Correct Four times as much, 1.8 W
Once the prediction is locked in, the Scene changes
Current
2 C
The Scene then shows: 1.8 W · I²R heating inside the cell/nodes/cell/outputs/batteryHeat
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Four times, 1.8 W: the heat goes with the square of the current.

Hint
The heat in a resistor is $I^2 R$ ⁠.

Worked solution
1. $P = I^2 R$ ⁠: at 3 A, $9 \times 0.05 = 0.45$ ⁠ watts.
2. At 6 A: $36 \times 0.05 = 1.8$ ⁠ watts, four times as much.
That is why fast charging heats a battery far more than slow charging, and why fast chargers slow down as the battery warms.
Step 10 of 10
Open this step
How full it is
Practice · Reach the target
Question
With no current flowing, a lithium battery’s voltage tells how charged it is. Lower the Charge level until the voltage at rest is 3.60 V, and press Check.

Goal
- 3.6 ± 0.01
  At the start of the step: 3.8
Setup, choices and feedback
The Scene at this step
Current
1 C
Charge level
60 %
Zoom to the ions
Cell
Read from the Scene at
Open-circuit voltage at this state of charge/nodes/cell/outputs/batteryRestVoltage
One setting that reaches it
Current
-2 C
Charge level
20 %
Feedback for a right answer
At 20%: from here the voltage drops fast, and the phone warns that the battery is low.

Hint
At 60% the voltage at rest is 3.80 V, and an empty battery reads 3.0 V.

Worked solution
1. As the battery empties, less lithium is left in the graphite, and the voltage at rest falls.
2. Between 20% and 80% it falls slowly, and fast at the ends.
3. 3.60 V comes at 20%.
That is how a phone estimates how much battery is left: it measures the voltage, and corrects it for the current flowing at that moment.

Across the unit · 14 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Oxidation and reduction

Losing electrons

Question

Correct answer

The Scene at this step

Choices

Worked solution

Where the electrons go

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Oxidation numbers

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Always together

Question

Correct answer

The Scene at this step

Choices

Worked solution

The reactivity series

Down the group

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Why

Question

Correct answer

The Scene at this step

Choices

Worked solution

Group 2

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Displacement

Question

Correct answer

The Scene at this step

Choices

Worked solution

Not the other way

Question

Correct answer

The Scene at this step

Choices

Worked solution

Metals and acids

Question

Correct answer

The Scene at this step

Choices

Worked solution

Galvanic cells

A cell’s voltage

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Choosing the metals

Question

Goal

The Scene at this step

One setting that reaches it

Hint