Electrode Potentials and Redox Reactions Revision Notes | A Level AQA Chemistry

1. Why Electrode Potentials Matter

Every time a metal corrodes, a battery discharges, or a fuel cell powers a vehicle, the driving force behind it is a difference in electrode potential. Understanding these potentials lets you predict which redox reactions will actually occur — and which ones will not.

At A Level, you need to go beyond recognising oxidation and reduction. You need to quantify the tendency of species to gain or lose electrons, compare those tendencies on a common scale, and use that scale to make predictions about real reactions. That scale is built from standard electrode potentials.

2. Half-Cells and the Need for a Reference Point

Picture a beaker containing $\text{Cu}^{2+}$(aq) ions and a copper metal electrode dipping into the solution. At the surface of the copper, an equilibrium is established: $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Cu}(\text{s})$. The position of this equilibrium determines whether the electrode is relatively positive or negative compared to the solution — and that difference in electrical potential is what we want to measure.

The problem is that you cannot measure a single half-cell potential in isolation. You always need two half-cells connected together to get a reading on a voltmeter. To make all measurements comparable, chemists agreed on one universal reference: the standard hydrogen electrode (SHE), assigned a potential of exactly 0.00 V.

The platinum electrode is inert — it does not participate in the reaction. Its role is purely to conduct electrons and provide a surface for the equilibrium to be established. When you connect any other half-cell to the SHE and measure the voltage under standard conditions, the reading you get is the standard electrode potential, $E^\circ$, of that half-cell.

3. Standard Conditions and Why They Are Non-Negotiable

The electrode potential of a half-cell is not a fixed, unchanging number — it shifts when you change the temperature, pressure, or concentration. This is a crucial point that the AQA specification emphasises: the standard electrode potential only applies under the defined standard conditions of 298 K, 100 kPa, and 1.00 mol dm$^{-3}$ ion concentration.

Consider the copper half-cell again. If you increase the concentration of $\text{Cu}^{2+}$(aq) above 1.00 mol dm$^{-3}$, the equilibrium $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Cu}(\text{s})$ shifts to the right. More reduction occurs, the electrode becomes more positive, and the measured potential rises above the standard value. The reverse is true at lower concentrations.

Temperature has a similar effect: higher temperatures alter the position of the equilibrium and therefore change the measured potential. This is why the Nernst equation (which quantifies these changes mathematically) exists — though at A Level you are not required to use it. What you are required to know is that non-standard conditions produce non-standard potentials, and that $E^\circ$ values are only directly comparable when all measurements are made under identical standard conditions.

4. The Electrochemical Series

Once you have measured the $E^\circ$ values of many half-cells against the SHE, you can arrange them in order of increasing potential. This ranked list is the electrochemical series. Each entry is written as a reduction half-equation — that is, with electrons on the left-hand side.

Reading the series gives you two immediate pieces of information. Species on the left-hand side of half-equations with more positive $E^\circ$ values are stronger oxidising agents — they have a greater tendency to accept electrons and be reduced. Species on the right-hand side of half-equations with more negative $E^\circ$ values are stronger reducing agents — they have a greater tendency to donate electrons and be oxidised.

A useful way to remember this: the more positive the $E^\circ$, the more the equilibrium lies to the right, favouring the reduced form. Chlorine ($E^\circ = +1.36$ V) is a far stronger oxidising agent than copper ($E^\circ = +0.34$ V), while lithium ($E^\circ = -3.04$ V) is a far stronger reducing agent than zinc ($E^\circ = -0.76$ V).

5. Using $E^\circ$ Values to Predict Reaction Direction

This is where the electrochemical series becomes a predictive tool. To determine whether a redox reaction is feasible under standard conditions, you combine two half-cells and calculate the overall cell EMF using the formula $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$, where the cathode is the more positive half-cell (reduction occurs there) and the anode is the more negative half-cell (oxidation occurs there).

The rule is straightforward: if $E^\circ_{\text{cell}}$ is positive, the reaction is thermodynamically feasible in the forward direction under standard conditions. If it is negative, the reaction is not feasible in that direction — but it would be feasible in the reverse direction.

There is an important caveat here that examiners test regularly. A positive $E^\circ_{\text{cell}}$ tells you the reaction is thermodynamically feasible — it does not guarantee the reaction will occur at a measurable rate. Kinetic barriers (high activation energy) can prevent a thermodynamically feasible reaction from proceeding at room temperature. Always distinguish between thermodynamic feasibility and kinetic viability in your exam answers.

Worked example: Predicting Feasibility: Fe$^{3+}$ oxidising I$^-$

Use $E^\circ$ values to predict whether $\text{Fe}^{3+}$(aq) will oxidise $\text{I}^-$(aq) under standard conditions. Given: $\text{Fe}^{3+}(\text{aq}) + e^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})$, $E^\circ = +0.77$ V; $\text{I}_2(\text{aq}) + 2e^- \rightleftharpoons 2\text{I}^-(\text{aq})$, $E^\circ = +0.54$ V.

1. Fe$^{3+}$(aq) gains an electron → Fe$^{2+}$(aq): this is reduction.\nI$^-$(aq) loses an electron → I$_2$(aq): this is oxidation. — The species with the more positive $E^\circ$ is reduced (cathode). Fe$^{3+}$/Fe$^{2+}$ has $E^\circ = +0.77$ V, which is more positive than I$_2$/I$^-$ at $+0.54$ V, so Fe$^{3+}$(aq) is reduced and I$^-$(aq) is oxidised.
2. Cathode (reduction): $\text{Fe}^{3+}(\text{aq}) + e^- \rightarrow \text{Fe}^{2+}(\text{aq})$, $E^\circ = +0.77$ V\nAnode (oxidation): $2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2e^-$, $E^\circ = +0.54$ V — At the anode, the half-equation is reversed from the standard reduction form. The $E^\circ$ value used in the formula is still the standard reduction potential — do not change its sign.
3. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$\n$E^\circ_{\text{cell}} = +0.77 - (+0.54)$\n$E^\circ_{\text{cell}} = +0.23$ V — The subtraction is straightforward here: $0.77 - 0.54 = 0.23$. Both values are positive, so take care with the arithmetic.
4. $E^\circ_{\text{cell}} = +0.23$ V (positive) — A positive $E^\circ_{\text{cell}}$ means the reaction is thermodynamically feasible under standard conditions. Fe$^{3+}$(aq) will oxidise I$^-$(aq) to I$_2$(aq) under standard conditions.

Answer: $E^\circ_{\text{cell}} = +0.23$ V. The reaction is feasible: Fe$^{3+}$(aq) oxidises I$^-$(aq) to I$_2$(aq) under standard conditions.

Notice the method: always identify which half-cell has the more positive $E^\circ$ (that one is reduced, at the cathode), then subtract the anode potential from the cathode potential. The sign of the result tells you the direction of feasibility.

6. A Second Worked Example: Predicting Non-Feasibility

It is equally important to recognise when a reaction is not feasible. Consider whether $\text{Cu}$(s) will reduce $\text{Fe}^{2+}$(aq) to $\text{Fe}$(s) under standard conditions. Given: $\text{Fe}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Fe}(\text{s})$, $E^\circ = -0.44$ V; $\text{Cu}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Cu}(\text{s})$, $E^\circ = +0.34$ V.

Worked example: Predicting Non-Feasibility: Cu reducing Fe$^{2+}$

Predict whether Cu(s) will reduce Fe$^{2+}$(aq) under standard conditions. $E^\circ$(Fe$^{2+}$/Fe) = $-0.44$ V; $E^\circ$(Cu$^{2+}$/Cu) = $+0.34$ V.

1. Proposed: Cu(s) reduces Fe$^{2+}$(aq) This means Cu is oxidised: $\text{Cu}(\text{s}) \rightarrow \text{Cu}^{2+}(\text{aq}) + 2e^-$ (anode) And Fe$^{2+}$ is reduced: $\text{Fe}^{2+}(\text{aq}) + 2e^- \rightarrow \text{Fe}(\text{s})$ (cathode) — We are testing a specific direction. Assign cathode and anode according to the proposed reaction, not by which $E^\circ$ is more positive.
2. $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ $E^\circ_{\text{cell}} = (-0.44) - (+0.34)$ $E^\circ_{\text{cell}} = -0.44 - 0.34 = -0.78$ V — Subtracting a positive number from a negative number: $-0.44 - 0.34 = -0.78$. The result is negative.
3. $E^\circ_{\text{cell}} = -0.78$ V (negative) — A negative $E^\circ_{\text{cell}}$ means the reaction as written is NOT feasible under standard conditions. The reverse reaction — Fe(s) reducing Cu$^{2+}$(aq) — would be feasible, with $E^\circ_{\text{cell}} = +0.78$ V.

Answer: $E^\circ_{\text{cell}} = -0.78$ V. Cu(s) will NOT reduce Fe$^{2+}$(aq) under standard conditions. The reaction is not thermodynamically feasible in this direction.

This example reinforces a key pattern: the spontaneous direction of a redox reaction always has the more positive half-cell acting as the cathode (reduction). If you try to force the less positive half-cell to act as the cathode, you get a negative $E^\circ_{\text{cell}}$ and the reaction is not feasible.

7. Measuring Standard Electrode Potentials in Practice

To measure the $E^\circ$ of, say, the $\text{Cu}^{2+}$/Cu half-cell, you connect it to the standard hydrogen electrode using a salt bridge and a high-resistance voltmeter. The salt bridge — typically a tube of saturated KCl(aq) in a gel — allows ions to flow between the two half-cells to complete the circuit without mixing the solutions. The high-resistance voltmeter ensures negligible current flows, so the system remains close to equilibrium and the reading reflects the true potential difference.

The salt bridge is essential: without it, no current can flow and the voltmeter reads zero. With it, the circuit is complete and the voltmeter displays the $E^\circ$ of the test half-cell directly, since the SHE contributes 0.00 V to the reading.

8. Exam Technique and Common Mistakes

AQA questions on electrode potentials tend to cluster around three areas: stating standard conditions precisely, calculating $E^\circ_{\text{cell}}$ correctly, and evaluating feasibility with appropriate caveats. Each of these has a characteristic trap.

Bridge: a positive E°cell tells you the reaction is thermodynamically feasible under standard conditions, but it says nothing on its own about whether the reaction will be fast enough to observe.

Taken together, the four specification requirements for this topic form a coherent framework: standard conditions define when $E^\circ$ values are valid; the SHE provides the reference point from which all values are measured; the electrochemical series organises those values into a predictive tool; and the cell EMF calculation applies that tool to determine whether a specific redox reaction is thermodynamically feasible.