# Standard electrode potentials

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u19-standard-electrode-potentials/

This sub-topic covers the definition, measurement and application of standard electrode potentials ($E^\circ$), a core concept for predicting redox reaction feasibility and calculating cell potential in all electrochemical cells for CIE A-Level Chemistry.

**Prerequisites:** [Redox reactions and half equations](https://www.owlsprep.com/study/cie-9701-u10-redox-reactions/); [Electrochemical cell basic structure](https://www.owlsprep.com/study/cie-9701-u19-electrochemical-cell-structure/)

## Learning objectives

- Define standard electrode potential and the standard hydrogen electrode reference
- Measure standard electrode potentials relative to the standard hydrogen electrode
- Calculate standard cell potentials and predict redox reaction feasibility
- Identify limitations of predictions made from standard electrode potentials

## Definitions and the Standard Hydrogen Electrode reference

**Standard electrode potential** — The electromotive force (e.m.f.) of a half-cell connected to the standard hydrogen electrode under standard conditions: 298 K, 1 mol dm⁻³ concentration of all aqueous ions, 1 atm pressure for all gaseous species.

*Notation:* $E^\circ$

*Example:* $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$

Half-cell potentials cannot be measured in isolation, because a half-reaction cannot occur on its own. All standard electrode potentials are measured relative to a fixed reference half-cell: the standard hydrogen electrode (SHE).

**Standard Hydrogen Electrode (SHE)** — A reference half-cell consisting of an inert platinum electrode immersed in 1 mol dm⁻³ $\text{H}^+$ ions, with hydrogen gas at 1 atm pressure bubbled continuously over the electrode, held at 298 K.

*Notation:* $E^\circ(\text{H}^+/\text{H}_2) = 0.00\ \text{V}$ (by definition)

*Example:* All other $E^\circ$ values are measured relative to the SHE, which is arbitrarily assigned an E° of 0.00 V.

**Worked example:** Describe how you would measure the standard electrode potential of the $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell.

1. Set up the standard hydrogen electrode as the reference half-cell, connect it to the $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell via a salt bridge (usually saturated $\text{KNO}_3$).
2. The $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell uses an inert platinum electrode, immersed in a solution with 1 mol dm⁻³ concentration of both $\text{Fe}^{3+}$ and $\text{Fe}^{2+}$ ions, held at 298 K.
3. Connect the two electrodes to a high-resistance voltmeter, which measures the e.m.f. of the full cell.
4. The voltmeter reading equals the standard electrode potential of the $\text{Fe}^{3+}/\text{Fe}^{2+}$ half-cell, since $E^\circ(\text{SHE}) = 0.00\ \text{V}$.

> **Exam tip:** Always list all three standard conditions when describing E° to earn full marks.

## Calculating standard cell potential $E^\circ_{\text{cell}}$

When two half-cells combine to form a full electrochemical cell, the overall standard cell potential can be calculated from the individual standard electrode potentials (all quoted as reduction potentials):

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode (reduction)}} - E^\circ_{\text{anode (oxidation)}}$$

The half-cell with the more negative $E^\circ$ will always undergo oxidation (is the anode), and the half-cell with the more positive $E^\circ$ will always undergo reduction (is the cathode).

**Worked example:** Calculate $E^\circ_{\text{cell}}$ for a cell made of $\text{Zn}^{2+}/\text{Zn}$ ($E^\circ = -0.76$ V) and $\text{Cu}^{2+}/\text{Cu}$ ($E^\circ = +0.34$ V).

1. Identify the more negative $E^\circ$: $\text{Zn}^{2+}/\text{Zn}$ is more negative, so oxidation occurs here (anode).
2. Assign roles: $\text{Cu}^{2+}/\text{Cu}$ is the cathode, where reduction occurs.
3. Substitute into the formula:
4. $$E^\circ_{\text{cell}} = (+0.34) - (-0.76) = +1.10\ \text{V}$$

**Check your understanding**

Test your understanding:

1. What is $E^\circ_{\text{cell}}$ for $\text{Ag}^+/\text{Ag}$ ($E^\circ = +0.80$ V) and $\text{Mg}^{2+}/\text{Mg}$ ($E^\circ = -2.37$ V)?

   - A: -1.57 V
   - B: +3.17 V
   - C: +1.57 V
   - D: -3.17 V

   *Why:* Mg has the more negative E°, so it oxidises. $E^\circ_{\text{cell}} = 0.80 - (-2.37) = +3.17$ V.

## Predicting feasibility of redox reactions

A redox reaction is thermodynamically feasible if the overall standard cell potential $E^\circ_{\text{cell}}$ is positive. To test feasibility of a proposed reaction, split it into two half-reactions, calculate $E^\circ_{\text{cell}}$, and check if it is positive.

**Worked example:** Predict whether bromine water will oxidise iron(II) ions to iron(III) ions. Given $E^\circ(\text{Br}_2/\text{Br}^-) = +1.07$ V, $E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77$ V.

1. Write the overall proposed reaction: $\text{Br}_2 + 2\text{Fe}^{2+} \rightarrow 2\text{Br}^- + 2\text{Fe}^{3+}$
2. Split into half-reactions: Reduction: $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$; Oxidation: $2\text{Fe}^{2+} \rightarrow 2\text{Fe}^{3+} + 2e^-$
3. Identify reduction and oxidation half-cells for the formula:
4. $$E^\circ_{\text{cell}} = E^\circ(\text{Br}_2/\text{Br}^-) - E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = 1.07 - 0.77 = +0.30\ \text{V}$$
5. Since $E^\circ_{\text{cell}}$ is positive, the reaction is thermodynamically feasible: bromine does oxidise iron(II) ions.

> **info**
>
> Feasibility only holds for standard conditions. If concentrations, temperature or pressure deviate from standard, the actual cell potential will change, which can reverse feasibility.

## Limitations of E° predictions

Even if $E^\circ_{\text{cell}}$ is positive, a reaction may not occur in practice. There are two key limitations to predictions from standard electrode potentials:

- **Kinetic effects**: $E^\circ$ only describes thermodynamic feasibility, it gives no information about activation energy. If activation energy is very high, the reaction will proceed too slowly to be observed, even if it is feasible.
- **Non-standard conditions**: $E^\circ$ is only defined for standard conditions. Deviations from 1 mol dm⁻³, 298 K or 1 atm will change the actual cell potential, which can make a feasible reaction non-feasible, or vice versa.

**Worked example:** Explain why solid manganese(IV) oxide reacts with concentrated hydrochloric acid, even though $E^\circ_{\text{cell}}$ is slightly negative under standard conditions.

1. Increasing HCl concentration increases $[\text{H}^+]$ and $[\text{Cl}^-]$, which shifts the equilibrium of the reaction to the right, following Le Chatelier's principle.
2. This changes the actual cell potential from negative (at 1 mol dm⁻³) to positive, making the reaction feasible, which is why this method is used to prepare chlorine gas in the laboratory.

## Common pitfalls

- **Wrong:** Forgetting to mention all three standard conditions when defining $E^\circ$.
  - Why it fails: CIE examiners require all three conditions to award full marks for definition questions.
  - Correct: Always list 298 K (25°C), 1 mol dm⁻³ ion concentration, and 1 atm pressure for gases.
- **Wrong:** Mixing up anode and cathode when calculating $E^\circ_{\text{cell}}$, leading to the wrong sign.
  - Why it fails: Many students swap the two values in the calculation, reversing the sign of the result.
  - Correct: Remember: more negative $E^\circ$ = oxidation at anode; $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$.
- **Wrong:** Changing the sign of $E^\circ$ when reversing a half-equation in calculations.
  - Why it fails: All $E^\circ$ values are quoted as reduction potentials, so the sign should not be changed when using the standard formula.
  - Correct: Leave $E^\circ$ values as their quoted reduction potentials, and just substitute into the $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ formula.
- **Wrong:** Claiming a positive $E^\circ_{\text{cell}}$ guarantees the reaction will occur at an observable rate.
  - Why it fails: $E^\circ$ only describes thermodynamic feasibility, not the kinetics of the reaction.
  - Correct: Always acknowledge that a high activation energy can prevent the reaction from occurring even if it is thermodynamically feasible.

## Cheatsheet

| Concept | Key Rule/Formula | Notes |
| --- | --- | --- |
| Standard $E^\circ$ definition | Measured vs SHE at 298 K, 1 mol dm⁻³, 1 atm | SHE $E^\circ = 0.00$ V by definition |
| $E^\circ_{\text{cell}}$ calculation | $E^\circ_{\text{cell}} = E^\circ_{\text{cathode (reduction)}} - E^\circ_{\text{anode (oxidation)}}$ | More negative $E^\circ$ = oxidation |
| Feasibility rule | Reaction is feasible if $E^\circ_{\text{cell}} > 0$ V | Only applies to standard conditions |
| Key limitations | 1. No kinetic information 2. Only for standard conditions | High activation energy stops slow reactions |

## What's next

Standard electrode potentials are the foundational concept for all further electrochemistry in CIE A-Level Chemistry, including calculating the Gibbs free energy change of redox reactions, predicting the products of electrolysis, and understanding the behaviour of commercial electrochemical cells like batteries and fuel cells. Questions on standard electrode potentials regularly appear in both paper 2 and paper 4, typically accounting for 8-12 marks per exam, so mastery of this sub-topic is critical for achieving high grades. The concepts here build directly on the redox chemistry you learned in AS Level, and lead to more advanced applications in the rest of unit 19.

- [Commercial Electrochemical Cells](https://www.owlsprep.com/study/cie-9701-u19-electrochemical-cells/)
- [Applications of electrode potentials](https://www.owlsprep.com/study/cie-9701-u19-applications-of-electrode-potentials/)
- [Further reaction kinetics](https://www.owlsprep.com/study/cie-9701-u20-overview/)

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