Redox and electrode potentials

In a nutshell

This subtopic takes the redox ideas from AS and puts numbers on them. You learn to build full redox equations from half-equations, use redox titrations to find unknown amounts, and use standard electrode potentials (E^⦵) to predict whether a redox reaction can happen.

The same ideas explain how batteries (storage cells) and fuel cells produce a voltage.

Assumed knowledge: Redox: oxidation numbers, oxidation and reduction; Acids: titration technique and calculations; Chemical equilibrium: le Chatelier's principle.

Core content

Redox: oxidising agents and reducing agents

Every redox reaction has a species that takes electrons and a species that gives them.

• An oxidising agent is a species that gains electrons from another species (an electron acceptor). The oxidising agent is itself reduced.
• A reducing agent is a species that loses electrons to another species (an electron donor). The reducing agent is itself oxidised.

Example: in 2Fe³⁺(aq) + 2I^-(aq) → 2Fe²⁺(aq) + I₂(aq)

• Fe³⁺ gains electrons (+3 → +2), so Fe³⁺ is the oxidising agent.
• I^- loses electrons (-1 → 0), so I^- is the reducing agent.

Always name the whole species that does the job: "manganate(VII) ions, MnO₄^-" is the oxidising agent, not "Mn" or "manganese". "Iodide ions" is the reducing agent, not "iodine".

Redox: constructing half-equations

A half-equation shows just the oxidation or just the reduction, with the electrons written in.

• In a reduction half-equation the electrons are on the left (gained): Cl₂ + 2e^- → 2Cl^-
• In an oxidation half-equation the electrons are on the right (lost): Fe²⁺ → Fe³⁺ + e^-

A half-equation must balance for atoms and for charge.

When oxygen atoms change (MnO₄^- → Mn²⁺, Cr₂O₇^2- → Cr³⁺), balance in acid with H₂O and H⁺. Follow these steps in order:

1. Balance the element that changes oxidation number.
2. Add electrons equal to the total change in oxidation number.
3. Add H₂O to balance the O atoms.
4. Add H⁺ to balance the H atoms.
5. Check the charge is the same on both sides.

For MnO₄^- → Mn²⁺ (Mn goes from +7 to +2, a change of 5):

Still don't get it? · balancing half-equations in acid

Think of a shop till. Every item that leaves the shop has to be paid for, and at the end of the day the till must balance exactly.

Now do it one small step at a time:

1. The "items" are the atoms. Whatever atoms go in must come out, so first make the element that is changing (Mn, Cr, S) the same on both sides.
2. The oxidation number change tells you how many electrons move. Mn going from +7 to +2 has gained 5 electrons, so write 5e^- on the left.
3. Oxygen atoms cannot just vanish. In acid solution, spare O atoms leave as water, so add H₂O to the side that is short of O.
4. Now that side has H atoms the other side does not. Acid solutions are full of H⁺, so add H⁺ to the other side.
5. Finally, check the "till": add up the charges on each side. If they match, the half-equation is right. If they do not, the electron count in step 2 is wrong.

Back to the exam: a half-equation earns its mark only if it is balanced for atoms and charge, with the electrons on the correct side.

Redox: combining half-equations into a full equation

To build the full redox equation, combine one reduction half-equation with one oxidation half-equation:

1. Multiply one or both half-equations so the numbers of electrons are equal.
2. Add the two half-equations together.
3. Cancel the electrons, and any H⁺ or H₂O that appears on both sides.

For MnO₄^- with Fe²⁺, the iron half-equation is multiplied by 5:

You can also balance a redox equation with oxidation numbers alone: the total increase in oxidation number must equal the total decrease. Here, 5 Fe each go up by 1 (+5 in total) and 1 Mn goes down by 5.

Redox: interpreting and predicting electron transfer

You will be given reactions you have never met, often with transition metals. For any of them you can:

• assign oxidation numbers to find what is oxidised and what is reduced
• identify the oxidising agent and reducing agent
• write the two half-equations, and combine them into the full equation
• use electrode potentials (later in this note) to predict whether the reaction happens.

For transition-element reactions you are not expected to recall the equations: you construct them from half-equations and oxidation numbers.

Redox titrations: manganate(VII) with iron(II)

Acidified potassium manganate(VII), KMnO₄, is used to find the amount of a reducing agent such as Fe²⁺.

MnO₄^-(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

The method:

1. Pipette a measured volume (e.g. 25.0 cm³) of the Fe²⁺ solution into a conical flask.
2. Add an excess of dilute sulfuric acid to supply the H⁺ ions.
3. Fill the burette with KMnO₄(aq). It is so deeply coloured that you read the top of the meniscus.
4. Add KMnO₄ while swirling. The purple colour disappears as MnO₄^- is reduced to almost colourless Mn²⁺.
5. The end point is the first permanent pale pink colour: the first drop of MnO₄^- that is left unreacted.

No indicator is needed: the titration is self-indicating, because MnO₄^- is purple and its product is colourless.

Why dilute sulfuric acid, and not another acid:

• Hydrochloric acid is unsuitable. MnO₄^- oxidises Cl^- to Cl₂ (E^⦵ MnO₄^-/Mn²⁺ = +1.51 V is more positive than E^⦵ Cl₂/Cl^- = +1.36 V). Extra manganate(VII) is used up, so the titre is too large.
• Too little acid means MnO₄^- is reduced only to brown MnO₂, which hides the end point and changes the reacting ratio.

Redox titrations: iodine with thiosulfate

Sodium thiosulfate, Na₂S₂O₃, reacts with iodine:

2S₂O₃^2-(aq) + I₂(aq) → 2I^-(aq) + S₄O₆^2-(aq)

This titration is used to find the amount of an oxidising agent, in two stages:

1. The oxidising agent reacts with an excess of potassium iodide, KI(aq). It oxidises I^- to I₂, so the solution turns yellow-brown.
2. The iodine formed is titrated with a standard solution of thiosulfate from the burette.

Following the colour:

The key ratio is 2 S₂O₃^2- : 1 I₂. Combine it with the ratio from stage 1 to link the thiosulfate to the original oxidising agent (Worked example 3).

Redox titrations: calculations

The technique (standard solution, rinsing, reading to the nearest 0.05 cm³, concordant titres within 0.10 cm³) is the same as for acid-base titrations in Acids.

Every redox titration calculation follows one chain:

1. Moles of titrant used: $n = c \times V$ (with V in dm³).
2. Ratio from the balanced equation (or equations) to get moles of the substance in the flask.
3. Scale up if the portion in the flask came from a larger volume (25.0 cm³ from 250.0 cm³ is × 10).
4. Convert to what is asked: concentration, mass, percentage purity, or the number of electrons transferred.

Non-familiar systems. You may meet a titration you have never seen, with the equation or half-equations given. If the equation is not given, use moles of electrons: the moles of electrons lost by the reducing agent equal the moles of electrons gained by the oxidising agent. This lets you find an unknown oxidation number (Worked example 4).

Electrode potentials: half-cells and cells

A half-cell contains the two species of a redox pair, for example a copper rod in a solution of Cu²⁺(aq). Each half-cell has an equilibrium written as a reduction:

Cu²⁺(aq) + 2e^- ⇌ Cu(s)

Joining two half-cells makes an electrochemical cell:

• The two electrodes are joined by a wire through a high-resistance voltmeter.
• The two solutions are joined by a salt bridge, usually filter paper soaked in saturated KNO₃(aq).

The salt bridge lets ions move between the solutions to complete the circuit. Electrons do not flow through it: they flow through the external wire.

The voltmeter has a high resistance so that almost no current flows. It then measures the maximum potential difference of the cell.

Electrode potentials: standard electrode potential and the hydrogen electrode

You cannot measure the potential of one half-cell on its own, only the difference between two. So every half-cell is measured against the same reference: the standard hydrogen electrode, given a potential of exactly 0 V.

2H⁺(aq) + 2e^- ⇌ H₂(g)    E^⦵ = 0.00 V

The standard hydrogen electrode is:

• hydrogen gas at 100 kPa, bubbled over
• a platinum electrode, dipping into
• a solution with [H⁺(aq)] = 1 mol dm^-3 (e.g. 1 mol dm^-3 HCl(aq))
• at 298 K.

The standard electrode potential, E^⦵, is the e.m.f. of a half-cell compared with a standard hydrogen half-cell, measured at 298 K, with solution concentrations of 1 mol dm^-3 and a gas pressure of 100 kPa.

Reading the sign:

• If the half-cell is the positive electrode compared with hydrogen, its E^⦵ is positive (Cu²⁺/Cu: +0.34 V).
• If it is the negative electrode, its E^⦵ is negative (Zn²⁺/Zn: -0.76 V).

E^⦵ values are always given in the exam; you do not need to learn them. The values used in this note are:

What the order tells you:

• More positive E^⦵: the species on the left gains electrons more readily, so it is a stronger oxidising agent (MnO₄^-, Cl₂).
• More negative E^⦵: the species on the right loses electrons more readily, so it is a stronger reducing agent (Mg, Zn).

Still don't get it? · what an electrode potential actually tells you

Picture a tug of war over electrons. Each half-cell is a team, and E^⦵ is how hard that team pulls electrons towards itself.

Now rebuild it step by step:

1. Every half-cell equilibrium is written the same way round, as a gain of electrons: Cu²⁺ + 2e^- ⇌ Cu.
2. You cannot measure one team's pull on its own, only who wins against someone else. So everyone is measured against the same opponent, the hydrogen electrode, which is called 0 V.
3. A positive E^⦵ means the half-cell pulls electrons harder than hydrogen. Its left-hand species (Cu²⁺) is good at grabbing electrons, so it is a good oxidising agent.
4. A negative E^⦵ means it pulls less hard than hydrogen. Its right-hand species (Zn) gives electrons away easily, so it is a good reducing agent.
5. Put any two half-cells together and the one with the more positive E^⦵ wins: it is reduced (goes left to right). The more negative one loses: it is oxidised (goes right to left). Electrons flow through the wire from the more negative electrode to the more positive one.

Back to the exam wording: "E^⦵ of Cu²⁺/Cu is more positive than E^⦵ of Zn²⁺/Zn, so Cu²⁺ is reduced and Zn is oxidised." Always compare the two values, and name which species is oxidised and which is reduced.

PAG8: Electrochemical cells

This practical covers how to set up cells and measure cell potentials for two kinds of half-cell.

1. A metal or non-metal in contact with its ions:

• Metal: the metal itself is the electrode, dipping into a solution of its ions, e.g. Zn(s) in 1.00 mol dm^-3 Zn²⁺(aq).
• Non-metal: a non-metal cannot be an electrode, so use platinum. For Cl₂/Cl^-, bubble Cl₂(g) at 100 kPa over a Pt electrode in 1.00 mol dm^-3 Cl^-(aq) (the hydrogen electrode is built the same way).

2. Ions of the same element in different oxidation states:

• Use a platinum electrode dipping into a solution containing both ions, e.g. Fe²⁺(aq) and Fe³⁺(aq).
• For a standard measurement, the two ions are either both 1 mol dm^-3 or equimolar (equal concentrations).
• Platinum is used because it is inert (it does not react) but conducts electrons to and from the ions.

Method:

1. Make up each half-cell solution at the required concentration using a volumetric flask.
2. Clean metal electrodes with emery paper (to remove any oxide layer), then rinse with distilled water.
3. Set up both half-cells, join the solutions with a freshly made salt bridge, and connect the electrodes to a high-resistance voltmeter.
4. Record the reading as soon as it is steady; the electrode connected to the positive terminal is the positive electrode.

Practical points:

• The salt bridge must not react with either solution. KCl would be unsuitable with Ag⁺(aq), for example, because it forms a precipitate of AgCl.
• A wire cannot replace the salt bridge: metal dipping into the solutions would set up extra electrode potentials of its own.

A typical PAG8 activity investigates the effect of concentration on cell potential. For a Zn/Cu cell with [Zn²⁺] kept at 1.00 mol dm^-3, diluting the Cu²⁺ solution lowers the cell potential:

Why: lowering [Cu²⁺] shifts Cu²⁺(aq) + 2e^- ⇌ Cu(s) to the left (le Chatelier), so the copper electrode potential becomes less positive and the cell potential falls.

Electrode potentials: calculating a standard cell potential

The standard cell potential is the difference between the two standard electrode potentials:

$E^{\ominus}_{\text{cell}} = E^{\ominus}(\text{positive electrode}) - E^{\ominus}(\text{negative electrode})$

The positive electrode is the half-cell with the more positive E^⦵. Done this way round, E^⦵_cell is always positive.

For the Zn/Cu cell:

$E^{\ominus}_{\text{cell}} = (+0.34) - (-0.76) = +1.10 \text{ V}$

When you combine half-equations into a full equation, you may multiply a half-equation to balance electrons, but never multiply its E^⦵ value. E^⦵ does not depend on how many moles you write.

Electrode potentials: predicting feasibility

To predict whether a redox reaction is feasible (can happen):

1. Write both half-equations as reductions, with their E^⦵ values.
2. The half-equation with the more positive E^⦵ goes forwards (reduction).
3. The half-equation with the more negative E^⦵ goes in reverse (oxidation).
4. The reaction in that direction is feasible, with a positive E^⦵_cell.

So the oxidising agent from the more positive half-cell reacts with the reducing agent from the more negative half-cell.

If a question asks about the reverse pairing (an oxidising agent from the more negative half-cell with a reducing agent from the more positive one), that reaction is not feasible.

Example: E^⦵ Cl₂/Cl^- (+1.36 V) is more positive than E^⦵ Br₂/Br^- (+1.09 V). So Cl₂ is reduced and Br^- is oxidised:

Cl₂(aq) + 2Br^-(aq) → 2Cl^-(aq) + Br₂(aq)    E^⦵_cell = +0.27 V

Bromine cannot oxidise chloride ions.

Electrode potentials: limitations of predictions

A positive E^⦵_cell says a reaction can happen. It does not guarantee you will see it.

1. Kinetics (rate). E^⦵ values say nothing about rate. A feasible reaction may have a large activation energy, so it is too slow to observe.

2. Concentration. E^⦵ values apply only to standard conditions (1 mol dm^-3). Real concentrations are usually different, which changes the electrode potentials:

• Increasing the concentration of the species on the left of a half-equation (e.g. Zn²⁺ in Zn²⁺ + 2e^- ⇌ Zn) shifts the equilibrium to the right, so E becomes more positive (less negative).
• Decreasing it shifts the equilibrium to the left, so E becomes less positive (more negative).

If the two E^⦵ values are close, a change in concentration can be enough to change the predicted outcome.

Storage and fuel cells: storage cells

A storage cell (battery) is an electrochemical cell sold as a portable energy source. The exam gives you its half-equations and E^⦵ values; you apply exactly the principles above.

With the data given, you can:

• decide which electrode is positive and which is negative (more positive E^⦵ = positive electrode)
• write the overall reaction on discharge, by combining the half-equations
• calculate the cell potential, E^⦵_cell
• explain recharging: an external supply forces the discharge reactions in reverse, re-forming the original reactants.

Modern rechargeable cells, such as lithium-based cells, are light and store a lot of energy. The benefits come with risks: toxic materials, and the risk of fire if a lithium cell is damaged or overheats.

Storage and fuel cells: fuel cells

A fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage.

The fuel and oxygen are supplied continuously from outside, so a fuel cell keeps working as long as they are fed in. It does not need recharging.

At each electrode:

• The fuel is oxidised at the negative electrode (it loses electrons).
• Oxygen is reduced at the positive electrode (it gains electrons).
• Electrons flow through the external circuit from the negative to the positive electrode.

For a hydrogen fuel cell with an alkaline electrolyte:

With the data E^⦵ = +0.40 V for O₂/OH^- and -0.83 V for H₂O/H₂:

$E^{\ominus}_{\text{cell}} = (+0.40) - (-0.83) = +1.23 \text{ V}$

With an acid electrolyte the electrode equations change (H₂ → 2H⁺ + 2e^- and O₂ + 4H⁺ + 4e^- → 2H₂O), but the overall reaction, and so the cell potential, is the same.

You do not need to recall fuel cell equations: the exam supplies them, with any E^⦵ data. What you must be able to do is explain what happens at each electrode.

Worked examples

1. Constructing and combining half-equations

(a) Write a half-equation for the reduction of acidified dichromate(VI) to chromium(III), then combine it with Fe²⁺ → Fe³⁺.

Cr goes from +6 to +3. There are 2 Cr atoms, so the total change is 6:

$\text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+}$ $\text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Charge check: left = -2 + 14 - 6 = +6; right = 2 × (+3) = +6.

Six electrons are gained, so the iron half-equation is multiplied by 6:

$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-$ $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$

Using a change of 3 electrons (forgetting there are two Cr atoms) is the commonest slip here.

(b) Combine MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O with SO₂ + 2H₂O → SO₄^2- + 4H⁺ + 2e^-.

The lowest common multiple of 5 and 2 is 10, so multiply by 2 and by 5:

$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$ $5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 20\text{H}^+ + 10e^-$

Add, then cancel 10e^-, 16H⁺ and 8H₂O from both sides:

$2\text{MnO}_4^- + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow 2\text{Mn}^{2+} + 5\text{SO}_4^{2-} + 4\text{H}^+$

Check: O = 8 + 10 + 2 = 20 on the left and 5 × 4 = 20 on the right; charge = -2 on both sides. Leaving H⁺ and H₂O uncancelled on both sides costs the mark.

2. Manganate(VII) titration: iron in a tablet

One iron tablet (containing iron(II) sulfate) was dissolved in dilute sulfuric acid and made up to 250.0 cm³. 25.0 cm³ portions were titrated with 0.00200 mol dm^-3 KMnO₄(aq).

Calculate the mass of iron in one tablet, in mg.

Mean of the concordant titres (1, 2 and 3 are all within 0.10 cm³; leave out the rough):

$\text{mean titre} = \frac{12.45 + 12.35 + 12.40}{3} = 12.40 \text{ cm}^3$

Moles of MnO₄^-:

$n(\text{MnO}_4^-) = 0.00200 \times \frac{12.40}{1000} = 2.48 \times 10^{-5} \text{ mol}$

Ratio MnO₄^- : Fe²⁺ = 1 : 5:

$n(\text{Fe}^{2+}) \text{ in } 25.0 \text{ cm}^3 = 5 \times 2.48 \times 10^{-5} = 1.24 \times 10^{-4} \text{ mol}$

Scale up to 250.0 cm³ (the whole tablet):

$n(\text{Fe}^{2+}) \text{ in } 250.0 \text{ cm}^3 = 10 \times 1.24 \times 10^{-4} = 1.24 \times 10^{-3} \text{ mol}$

Mass, using A_r(Fe) = 55.8:

$m = 1.24 \times 10^{-3} \times 55.8 = 0.0692 \text{ g} = 69.2 \text{ mg}$

Answer: 69.2 mg (3 s.f., matching 0.00200 and 25.0).

Where marks go: dividing by 5 instead of multiplying gives 2.77 mg; forgetting the × 10 gives 6.92 mg; including the rough titre in the mean gives the wrong titre.

3. Iodine-thiosulfate titration: chlorate(I) in bleach

10.0 cm³ of bleach was diluted to 250.0 cm³. A 25.0 cm³ portion was added to excess acidified KI(aq). The iodine released needed 18.20 cm³ of 0.100 mol dm^-3 Na₂S₂O₃(aq). Calculate the concentration of ClO^- in the original bleach.

ClO^-(aq) + 2I^-(aq) + 2H⁺(aq) → Cl^-(aq) + I₂(aq) + H₂O(l)

2S₂O₃^2-(aq) + I₂(aq) → 2I^-(aq) + S₄O₆^2-(aq)

Moles of thiosulfate:

$n(\text{S}_2\text{O}_3^{2-}) = 0.100 \times \frac{18.20}{1000} = 1.82 \times 10^{-3} \text{ mol}$

Ratio S₂O₃^2- : I₂ = 2 : 1:

$n(\text{I}_2) = \frac{1.82 \times 10^{-3}}{2} = 9.10 \times 10^{-4} \text{ mol}$

Ratio I₂ : ClO^- = 1 : 1, so in the 25.0 cm³ portion:

$n(\text{ClO}^-) = 9.10 \times 10^{-4} \text{ mol}$

Scale up to 250.0 cm³, which all came from the 10.0 cm³ of bleach:

$n(\text{ClO}^-) \text{ in } 250.0 \text{ cm}^3 = 10 \times 9.10 \times 10^{-4} = 9.10 \times 10^{-3} \text{ mol}$ $c = \frac{9.10 \times 10^{-3}}{10.0 / 1000} = 0.910 \text{ mol dm}^{-3}$

Answer: 0.910 mol dm^-3 (3 s.f.).

Where marks go: dividing by the 250.0 cm³ (giving 0.0364 mol dm^-3) instead of the 10.0 cm³ of bleach, or using a 1 : 1 ratio for thiosulfate : iodine (giving 1.82 mol dm^-3).

4. A non-familiar system: finding an oxidation number

25.0 cm³ of a 0.0500 mol dm^-3 solution of a metal ion X²⁺ was titrated with acidified 0.0200 mol dm^-3 KMnO₄(aq). The titre was 37.50 cm³. MnO₄^- was reduced to Mn²⁺. Find the oxidation number of X after the reaction.

$n(\text{X}^{2+}) = 0.0500 \times \frac{25.0}{1000} = 1.25 \times 10^{-3} \text{ mol}$ $n(\text{MnO}_4^-) = 0.0200 \times \frac{37.50}{1000} = 7.50 \times 10^{-4} \text{ mol}$

Each MnO₄^- gains 5 electrons (+7 → +2):

$n(e^-) \text{ gained} = 5 \times 7.50 \times 10^{-4} = 3.75 \times 10^{-3} \text{ mol}$

Electrons lost by X = electrons gained by MnO₄^-:

$\text{electrons lost per X} = \frac{3.75 \times 10^{-3}}{1.25 \times 10^{-3}} = 3$ $\text{new oxidation number} = +2 + 3 = +5$

Answer: +5. Each X lost 3 electrons, so its oxidation number rose by 3 from the starting value of +2. Giving "+3" (the change, not the final value) loses the mark.

5. Calculating a standard cell potential

A cell is made from a Mg²⁺/Mg half-cell (E^⦵ = -2.37 V) and an Ag⁺/Ag half-cell (E^⦵ = +0.80 V). Find E^⦵_cell and write the overall cell reaction.

Ag⁺/Ag is more positive, so silver is the positive electrode and is reduced. Mg is the negative electrode and is oxidised.

$E^{\ominus}_{\text{cell}} = (+0.80) - (-2.37) = +3.17 \text{ V}$

Balance electrons (Mg loses 2, each Ag⁺ gains 1):

Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s)

Answer: E^⦵_cell = +3.17 V. Doubling the silver value because of the "2Ag⁺" gives 3.97 V, which is wrong: never multiply E^⦵. Electrons flow in the wire from the Mg electrode to the Ag electrode.

6. Predicting feasibility

Use the E^⦵ values below to predict whether Fe³⁺(aq) will oxidise (a) I^-(aq) and (b) Br^-(aq).

(a) E^⦵ Fe³⁺/Fe²⁺ (+0.77 V) is more positive than E^⦵ I₂/I^- (+0.54 V). So Fe³⁺ is reduced and I^- is oxidised: feasible.

$E^{\ominus}_{\text{cell}} = (+0.77) - (+0.54) = +0.23 \text{ V}$

2Fe³⁺(aq) + 2I^-(aq) → 2Fe²⁺(aq) + I₂(aq)

(b) E^⦵ Fe³⁺/Fe²⁺ (+0.77 V) is less positive than E^⦵ Br₂/Br^- (+1.09 V). Fe³⁺ cannot oxidise Br^-: not feasible. (The reverse is feasible: Br₂ oxidises Fe²⁺, with E^⦵_cell = +0.32 V.)

Even for (a), the prediction is only that the reaction can happen under standard conditions. It might still be too slow (large activation energy), and non-standard concentrations change the E values.

7. A storage cell from given data

A lead-acid cell uses these half-cells:

(a) Calculate E^⦵_cell.

$E^{\ominus}_{\text{cell}} = (+1.69) - (-0.36) = +2.05 \text{ V}$

(b) Write the overall equation for discharge.

The PbO₂ half-cell (more positive) goes forwards; the Pb half-cell goes in reverse. Both involve 2 electrons, so just add and cancel:

Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄^2-(aq) → 2PbSO₄(s) + 2H₂O(l)

(c) How many cells in series are needed for a battery of about 12 V?

$\frac{12}{2.05} = 5.9 \;\Rightarrow\; 6 \text{ cells}, \quad 6 \times 2.05 = 12.3 \text{ V}$

On recharging, an external supply drives this equation in reverse, turning PbSO₄ back into Pb and PbO₂.

Common exam mistakes

• Naming an element as the oxidising agent. Students write "Mn" or "manganese"; the oxidising agent is MnO₄^- (manganate(VII) ions). Likewise the reducing agent is iodide ions, not "iodine".
• Mixing up "oxidised" and "oxidising agent". The oxidising agent is the species that is reduced.
• Electrons on the wrong side of a half-equation, or a half-equation that balances atoms but not charge. Always add up the charges on each side.
• Leaving electrons, H⁺ or H₂O on both sides of a combined equation. Cancel them.
• Counting one Cr, not two, in Cr₂O₇^2- → 2Cr³⁺: the half-equation needs 6e^-, not 3e^-.
• Adding an indicator to a manganate(VII) titration. It is self-indicating; the end point is the first permanent pale pink, not "colourless" and not "purple to colourless".
• Reading the bottom of the meniscus for KMnO₄(aq). It is too dark; read the top.
• Saying any acid will do for manganate(VII) titrations. It must be dilute sulfuric acid; HCl is oxidised to Cl₂, giving a titre that is too large.
• Adding starch at the start of an iodine-thiosulfate titration. Add it near the end point (pale yellow); the end point is blue-black to colourless ("clear" is not a colour).
• Using 1 : 1 for thiosulfate : iodine. It is 2 S₂O₃^2- : 1 I₂.
• Forgetting the 1 : 5 ratio for MnO₄^- : Fe²⁺, or forgetting to scale up from the 25.0 cm³ portion to the full volume.
• Leaving out a standard condition in the definition of E^⦵. The mark needs compared with the standard hydrogen half-cell, 298 K, 1 mol dm^-3 and 100 kPa. A vague phrase such as "room temperature and pressure" does not state the conditions.
• Getting the hydrogen electrode wrong: missing the platinum electrode, or writing H₂ at "1 mol dm^-3" (it is H⁺ at 1 mol dm^-3 and H₂ at 100 kPa).
• Saying electrons flow through the salt bridge. Ions move through the salt bridge; electrons flow through the wire.
• Multiplying E^⦵ when a half-equation is multiplied to balance electrons. E^⦵ stays the same.
• Subtracting the wrong way round and quoting a negative E^⦵_cell for a feasible reaction. Use E^⦵(positive) - E^⦵(negative).
• Saying a positive E^⦵_cell means the reaction "will happen quickly". E^⦵ says nothing about rate; a large activation energy can make it too slow.
• Getting the concentration effect backwards. Increasing the concentration of the species on the left of the half-equation makes E more positive.
• Saying a fuel cell "burns" the fuel or "stores" energy and needs recharging. It converts the energy of the fuel-oxygen reaction into a voltage, with fuel and oxygen supplied continuously.

Key definitions

• Oxidising agent: a species that gains electrons from another species (an electron acceptor), and is itself reduced.
• Reducing agent: a species that loses electrons to another species (an electron donor), and is itself oxidised.
• Half-equation: an equation showing either the oxidation or the reduction part of a redox reaction, including the electrons lost or gained.
• Standard electrode potential, E^⦵: the e.m.f. of a half-cell compared with a standard hydrogen half-cell, measured at 298 K with solution concentrations of 1 mol dm^-3 and a gas pressure of 100 kPa.
• Standard hydrogen electrode: hydrogen gas at 100 kPa bubbled over a platinum electrode in a solution with [H⁺(aq)] = 1 mol dm^-3 at 298 K; its potential is defined as 0 V.
• Standard cell potential, E^⦵_cell: the difference between the standard electrode potentials of the two half-cells: E^⦵(positive electrode) - E^⦵(negative electrode).
• Salt bridge: a connection between the two half-cell solutions that allows ions to move, completing the circuit (usually filter paper soaked in saturated KNO₃(aq)).
• Fuel cell: a cell that uses the energy from the reaction of a fuel with oxygen to create a voltage.

Specification

• I can explain and use the terms oxidising agent and reducing agent. (5.2.3 a)
• I can construct redox equations using half-equations and oxidation numbers. (5.2.3 b)
• I can interpret and predict reactions involving electron transfer. (5.2.3 c)
• I can describe the techniques and procedures for redox titrations, including Fe²⁺/MnO₄^- and I₂/S₂O₃^2-. (5.2.3 d)
• I can do structured and non-structured titration calculations from the results of Fe²⁺/MnO₄^- and I₂/S₂O₃^2- titrations. (5.2.3 e i)
• I can do titration calculations for redox systems I have not met before. (5.2.3 e ii)
• I can use the term standard electrode potential, E^⦵, and describe how it is measured using a hydrogen electrode. (5.2.3 f)
• I can describe how to measure the cell potential of a metal or non-metal in contact with its ions. (5.2.3 g i, PAG8)
• I can describe how to measure the cell potential of ions of the same element in different oxidation states, using a Pt electrode. (5.2.3 g ii, PAG8)
• I can calculate a standard cell potential by combining two standard electrode potentials. (5.2.3 h)
• I can predict whether a reaction is feasible from standard cell potentials, and explain the limitations of these predictions in terms of kinetics and concentration. (5.2.3 i)
• I can apply electrode potential principles to modern storage cells, using data I am given. (5.2.3 j)
• I can explain that a fuel cell uses the energy from the reaction of a fuel with oxygen to create a voltage, and describe the changes at each electrode. (5.2.3 k)

Related notes

• Redox: oxidation numbers, oxidation and reduction (the AS foundation for this note)
• Acids: titration technique and acid-base titration calculations
• Amount of substance: n = c × V, used in every titration calculation
• Chemical equilibrium: le Chatelier's principle, behind the effect of concentration on E
• Enthalpy and entropy: feasibility from free energy, and its own kinetic limitation
• Transition elements: redox reactions of iron, chromium and copper ions, built with the half-equations from this note