Define standard electrode potential, calculate standard cell potential and use it to predict the feasibility of redox reactions, describe the effect of concentration qualitatively, and apply Faraday's laws to electrolysis calculations

What this dot point is asking

SEAB wants you to define standard electrode potential against the standard hydrogen electrode, calculate standard cell potential and use its sign to predict whether a redox reaction is feasible, describe qualitatively how changing concentration shifts a potential, and apply Faraday's laws to electrolysis. Cell-potential feasibility and electrolysis mass or volume calculations are dependable Paper 2 questions.

The answer

Standard electrode potential

The standard electrode potential $E^{\circ}$ of a half-cell is its potential measured under standard conditions ($298$ K, $1$ bar, $1$ mol dm$^{-3}$ ion concentration) relative to the standard hydrogen electrode, which is defined as $0.00$ V. A more positive $E^{\circ}$ means the species is more readily reduced (a stronger oxidising agent).

Standard cell potential

When two half-cells are connected, the cell potential is:

The more positive electrode is the cathode (reduction, positive terminal). $E^{\circ}_{cell}$ is always positive for a spontaneous cell.

Predicting feasibility

A redox reaction is feasible under standard conditions if the cell potential built from its half-reactions is positive ($E^{\circ}_{cell} > 0$). Combine the reduction half-equation of the oxidising agent with the reverse (oxidation) of the other:

• $E^{\circ}_{cell} > 0$: reaction is thermodynamically feasible.
• $E^{\circ}_{cell} < 0$: not feasible as written (the reverse is feasible).

As with $\Delta G$, feasible does not mean fast; a positive $E^{\circ}_{cell}$ says nothing about rate.

The effect of concentration

Changing concentrations shifts an electrode potential (qualitatively, by Le Chatelier on the half-equation):

• Increasing the concentration of the oxidised species (the reactant being reduced) makes $E$ more positive.
• Increasing the concentration of the reduced species (the product) makes $E$ less positive.

A large enough concentration change can even reverse the feasibility of a borderline reaction.

Faraday's laws and electrolysis

In electrolysis, the amount of product is governed by the charge passed. The key relationships:

The half-equation gives the ratio of electrons to product. The routine:

1. Charge $Q = It$ (current in A, time in s).
2. Moles of electrons $= Q/F$.
3. Use the half-equation ratio to get moles of product.
4. Convert to mass ($\times M$) or gas volume ($\times 24.0$ dm cubed at r.t.p.).

Examples in context

Example 1. Why a sacrificial anode protects steel. Attaching a block of zinc ($E^{\circ} = -0.76$ V) to an iron structure ($E^{\circ} = -0.44$ V) makes the zinc the anode because it has the more negative potential, so zinc is oxidised in preference and the iron is protected. SEAB tests this by asking candidates to identify which metal corrodes from electrode potentials.

Example 2. Extracting reactive metals by electrolysis. Aluminium cannot be extracted by carbon reduction because it is too reactive, so it is obtained by electrolysis of molten aluminium oxide. A Faraday's-law calculation links the industrial current and time to the mass of aluminium produced, exactly the kind of quantitative electrolysis question that appears in Paper 2.

Try this

Q1. Define standard electrode potential. [2 marks]

• Cue. The potential of a half-cell relative to the standard hydrogen electrode under standard conditions ($298$ K, $1$ bar, $1$ mol dm$^{-3}$).

Q2. Given $E^{\circ}(\text{Ag}^+/\text{Ag}) = +0.80$ V and $E^{\circ}(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V, calculate $E^{\circ}_{cell}$ and state which metal dissolves. [3 marks]

• Cue. $E^{\circ}_{cell} = 0.80 - 0.34 = +0.46$ V; copper (more negative) is the anode and dissolves.

Q3. A current of $2.00$ A flows for $30.0$ minutes through aqueous copper(II) sulfate. Calculate the mass of copper deposited. ($F = 96500$, Ar(Cu) = $63.5$.) [3 marks]

• Cue. $Q = 2.00 \times 1800 = 3600$ C; $n(e^-) = 0.0373$; $\text{Cu}^{2+} + 2e^-$, so $n(\text{Cu}) = 0.01865$; mass $= 1.18$ g.