A Level Chemistry: Electrochemistry — Revision Notes

Condensed for the final weeks. For the full explanation, use the Electrochemistry study guide.

Standard conditions

298 K,  1 mol dm^-3 solutions,  100 kPa (or 1 atm) for gases
Reference: standard hydrogen electrode, E = 0.00 V by DEFINITION

Every E⦵ value is a measurement relative to the hydrogen electrode, which is assigned zero — it is not an absolute quantity.

Reading E⦵ values

Standard electrode potentials are written as reduction half-equations:

Cu2+ + 2e-  <=>  Cu        E = +0.34 V
Zn2+ + 2e-  <=>  Zn        E = -0.76 V

• More positive E⦵ → more readily reduced → stronger oxidising agent (left-hand species).
• More negative E⦵ → more readily oxidised → stronger reducing agent (right-hand species).

Cell e.m.f.

E_cell = E(more positive)  -  E(more negative)

For a cell freely built from two half-cells (as above), E_cell is always positive — you simply take the more positive electrode potential minus the more negative one.

The more negative electrode is the anode (oxidation, electrons released). The more positive is the cathode (reduction). Electrons flow through the external circuit from negative to positive.

Mnemonic that survives pressure: an ox, red cat — anode oxidation, reduction cathode.

Zn | Zn2+ || Cu2+ | Cu        E_cell = 0.34 - (-0.76) = +1.10 V

Single line = phase boundary, double line = salt bridge.

This is not the same calculation as testing feasibility of a specified reaction — see below. Only use “more positive minus more negative” when you are free to choose which half-cell is reduced and which is oxidised.

Feasibility

For a specified reaction (one species given as being reduced, another as being oxidised), calculate:

E_cell = E(species reduced)  -  E(species oxidised)

This can come out negative — and a negative value means the reaction is not feasible as written.

Worked example (infeasible): does Fe3+ oxidise Cl- to Cl2? Fe3+/Fe2+, E = +0.77 V (reduced); Cl2/Cl-, E = +1.36 V (oxidised).

E_cell = 0.77 - 1.36 = -0.59 V   ->  NOT feasible

Iron(III) is too weak an oxidising agent to oxidise chloride ions; the reverse reaction (Cl2 oxidising Fe2+) is the feasible one.

delta-G = -n F E_cell        F = 96500 C mol^-1

E_cell positive → ΔG negative → thermodynamically feasible.

Two essential caveats:

1. Feasible does not mean fast. A reaction with a large positive E_cell may still be immeasurably slow if the activation energy is high.
2. E⦵ values apply at standard conditions only. Changing concentration shifts the electrode potential — by Le Chatelier, increasing the concentration of the species on the left of a reduction half-equation makes E more positive.

Electrolysis

Electrode	Process	Charge
Cathode	Reduction — cations gain electrons	Negative
Anode	Oxidation — anions lose electrons	Positive

Note the reversal from an electrochemical cell: in electrolysis the cathode is negative; in a galvanic cell it is positive. The constant is that reduction always happens at the cathode.

Q = I t          Q = n F

To find mass deposited: Q = It, then n(e⁻) = Q/F, then divide by the number of electrons in the half-equation, then m = nM.

Selective discharge in aqueous solution

• Cathode: the less reactive cation (more positive E⦵) is discharged. Metals below hydrogen deposit; metals above it leave H₂ evolved instead.
• Anode: oxygen is released from hydroxide/water by default (hydroxide/water is easier to oxidise than most halides on electrode-potential grounds) unless a halide is present at high concentration, in which case the halogen is discharged instead — concentrated NaCl gives Cl₂, very dilute NaCl gives O₂. Fluoride is never discharged from aqueous solution at any concentration (F⁻/F₂, E⦵ = +2.87 V, far too positive).

The Nernst equation

Electrode potential is not fixed — it varies with the concentration of the aqueous ions involved:

E = E-standard + (0.059 / z) log([oxidised species] / [reduced species])

where z is the number of electrons transferred. Qualitatively, increasing the concentration of the oxidised species (on the left of the reduction half-equation) makes E more positive; increasing the concentration of the reduced species makes E less positive — consistent with Le Chatelier’s principle applied to the half-equilibrium.

Worked example: electrolysis

A current of 2.00 A is passed through aqueous CuSO₄ with copper electrodes for 3860 s. Calculate the mass of copper deposited. (F = 96,500 C mol⁻¹, A_r(Cu) = 63.5)

Q = It = 2.00 x 3860 = 7720 C
moles of electrons = Q / F = 7720 / 96500 = 0.0800 mol
Cu2+ + 2e- -> Cu, so moles of Cu = 0.0800 / 2 = 0.0400 mol
mass of Cu = 0.0400 x 63.5 = 2.54 g

Exam traps

• Subtracting the wrong way and reporting a negative e.m.f. for a spontaneous cell.
• Confusing anode polarity between electrolytic and galvanic cells.
• Forgetting that ΔG = −nFE_cell needs n = moles of electrons transferred, taken from the balanced overall equation.
• Treating “feasible” as “will happen quickly”.
• Applying E⦵ values to non-standard concentrations without comment.

Self-test

1. What is the standard hydrogen electrode’s potential, and why?
2. In Zn|Zn²⁺||Cu²⁺|Cu, which electrode is the anode and which way do electrons flow?
3. Calculate E_cell for Zn/Cu.
4. State the relationship between ΔG and E_cell, and what makes a reaction feasible.
5. Why is the cathode negative in electrolysis but positive in a galvanic cell?
6. A current of 2.00 A is passed through aqueous CuSO₄ with copper electrodes for 3860 s. Calculate the mass of copper deposited (F = 96,500 C mol⁻¹, A_r(Cu) = 63.5).
7. According to the Nernst equation, what happens to E if the concentration of the oxidised species increases?

Answers: 1. 0.00 V, by definition — it is the arbitrary reference against which all other electrode potentials are measured. 2. Zinc is the anode (more negative, oxidised); electrons flow externally from zinc to copper. 3. +0.34 − (−0.76) = +1.10 V. 4. ΔG = −nFE_cell; a positive E_cell gives a negative ΔG, so the reaction is thermodynamically feasible. 5. In electrolysis an external supply pushes electrons onto the cathode to drive reduction; in a galvanic cell reduction draws electrons in, making the cathode the positive terminal. Reduction occurs at the cathode in both cases. 6. Q = It = 7720 C; moles of electrons = 7720 ÷ 96500 = 0.0800 mol; moles of Cu = 0.0800 ÷ 2 = 0.0400 mol; mass = 0.0400 × 63.5 = 2.54 g. 7. E becomes more positive, since increasing the concentration of the oxidised species shifts the half-equilibrium towards reduction, consistent with Le Chatelier’s principle.