# Applications of electrode potentials

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

This subtopic explains how to use standard electrode potential ($E^\circ$) values to calculate cell potentials, predict redox reaction spontaneity, and identify electrolysis products. It also covers key limitations of these predictions.

**Prerequisites:** [Standard electrode potentials](https://www.owlsprep.com/study/cie-9701-u19-standard-electrode-potentials/); [Redox reactions and half-equations](https://www.owlsprep.com/study/cie-9701-u5-redox-reactions/)

## Learning objectives

- Calculate standard cell potential from standard electrode potential values
- Predict the spontaneity of redox reactions using E° values
- Predict products of electrolysis using electrode potential comparisons
- Explain limitations of predictions based on standard electrode potentials

## Calculating Standard Cell Potential

**Standard Cell Potential** — The potential difference between the cathode (reduction) and anode (oxidation) of an electrochemical cell under standard conditions.

*Notation:* $E^\circ_{cell}$

*Example:* For a Zn/Cu cell, $E^\circ_{cell} = E^\circ(Cu^{2+}/Cu) - E^\circ(Zn^{2+}/Zn)$

To calculate $E^\circ_{cell}$, follow the convention that reduction always occurs at the cathode, and oxidation always occurs at the anode. The core formula is:

$$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$$

**Worked example:** Given $E^\circ(Fe^{3+}/Fe^{2+}) = +0.77\ V$ and $E^\circ(Br_2/Br^-) = +1.07\ V$, calculate $E^\circ_{cell}$ for the reaction where $Fe^{2+}$ is oxidised by $Br_2$.

1. Step 1: Identify oxidation and reduction half-equations. $Fe^{2+}$ is oxidised (anode) and $Br_2$ is reduced (cathode).
2. Step 2: Substitute values into the formula:
3. $$E^\circ_{cell} = +1.07 - (+0.77) = +0.30\ V$$

> **Exam tip:** Always confirm which half-cell is the anode and which is the cathode before subtracting, to avoid sign errors.

## Predicting Redox Reaction Spontaneity

There is a direct relationship between $E^\circ_{cell}$ and the standard Gibbs free energy change $\Delta G^\circ$, which tells us if a reaction is spontaneous:

$$\Delta G^\circ = -nFE^\circ_{cell}$$

Where $n$ is the moles of electrons transferred and $F$ is Faraday's constant. From this relationship, we get two simple rules:

1. If $E^\circ_{cell} > 0$, $\Delta G^\circ < 0$: the forward reaction is spontaneous under standard conditions
2. If $E^\circ_{cell} < 0$, $\Delta G^\circ > 0$: the forward reaction is non-spontaneous, reverse reaction is spontaneous

**Worked example:** Predict if magnesium metal will displace lead ions from solution, given $E^\circ(Mg^{2+}/Mg) = -2.37\ V$ and $E^\circ(Pb^{2+}/Pb) = -0.13\ V$.

1. Step 1: Write the expected reaction: $Mg(s) + Pb^{2+}(aq) \rightarrow Mg^{2+}(aq) + Pb(s)$
2. Step 2: Identify oxidation (Mg, anode: $E^\circ = -2.37\ V$) and reduction (Pb²+, cathode: $E^\circ = -0.13\ V$)
3. Step 3: Calculate $E^\circ_{cell}$:
4. $$E^\circ_{cell} = -0.13 - (-2.37) = +2.24\ V$$
5. Step 4: Conclusion: $E^\circ_{cell}$ is positive, so the reaction is spontaneous, magnesium displaces lead.

**Check your understanding**

Test your understanding:

1. What does $E^\circ_{cell} = -0.45\ V$ mean for the forward reaction?

   - Forward reaction is spontaneous under standard conditions
   - Reverse reaction is spontaneous under standard conditions
   - No reaction can ever occur
   - Reaction will be spontaneous at higher temperature

   *Answer:* Reverse reaction is spontaneous under standard conditions

   *Why:* Negative $E^\circ_{cell}$ gives positive $\Delta G^\circ$, so forward is non-spontaneous, reverse is spontaneous.

## Predicting Products of Electrolysis

For electrolysis with inert electrodes, we can use $E^\circ$ values to predict which species is discharged at each electrode, for similar concentrations of ions:

- **Cathode (negative electrode, reduction):** The species with the *most positive* $E^\circ$ is reduced preferentially
- **Anode (positive electrode, oxidation):** The species with the *most negative* $E^\circ$ is oxidised preferentially

> **Common Mistake Alert**
>
> Always include water (present in all aqueous solutions) when comparing $E^\circ$ values. Overpotential and concentration effects can change outcomes for concentrated solutions.

**Worked example:** Predict the product at the cathode during electrolysis of dilute aqueous silver nitrate, given $E^\circ(Ag^+/Ag) = +0.80\ V$ and $E^\circ(2H^+/H_2) = 0.00\ V$.

1. Step 1: List all cations available for reduction at the cathode: $Ag^+(aq)$ and $H^+(aq)$ from water.
2. Step 2: Compare $E^\circ$ values: $Ag^+/Ag$ (+0.80 V) is more positive than $H^+/H_2$ (0.00 V).
3. Step 3: Conclusion: Silver ions are preferentially reduced, so solid silver metal is deposited at the cathode.

## Limitations of E° Predictions

Predictions based on standard $E^\circ$ values can fail for three key reasons:

- Non-standard conditions: $E^\circ$ is only valid for 1 mol dm⁻³ concentration, 298 K and 1 atm pressure. Deviations change the actual potential.
- Kinetics: $E^\circ$ only describes thermodynamics (spontaneity), not rate. A spontaneous reaction may have very high activation energy and proceed too slowly to observe.
- Overpotential: Extra voltage is required for gas discharge at electrodes, which can change the product of electrolysis.

**Worked example:** Explain why chlorine is produced at the anode during electrolysis of concentrated brine (NaCl), even though $E^\circ(O_2/H_2O/OH^-) = +0.40\ V$ and $E^\circ(Cl_2/Cl^-) = +1.36\ V$.

1. Based on $E^\circ$ values, OH- (with more negative $E^\circ$) should be oxidised preferentially to oxygen.
2. In concentrated brine, chloride ion concentration is much higher than 1 mol dm⁻³, and there is a high overpotential for oxygen discharge.
3. These effects make chloride oxidation to chlorine favourable, despite the standard $E^\circ$ prediction.

## Common pitfalls

- **Wrong:** Calculating $E^\circ_{cell}$ as $E^\circ_{anode} - E^\circ_{cathode}$ (reversed order of subtraction)
  - Why it fails: This gives the wrong sign for $E^\circ_{cell}$, leading to incorrect spontaneity predictions
  - Correct: Always use $E^\circ_{cell} = E^\circ(\text{cathode, reduction}) - E^\circ(\text{anode, oxidation})$
- **Wrong:** Assuming a positive $E^\circ_{cell}$ means the reaction will occur at an observable rate
  - Why it fails: $E^\circ$ only describes thermodynamic spontaneity, not the kinetics (rate) of reaction
  - Correct: Remember that high activation energy can make a spontaneous reaction too slow to observe
- **Wrong:** Forgetting to include water when predicting products of aqueous electrolysis
  - Why it fails: Water can be oxidised or reduced, so it must always be included in $E^\circ$ comparisons
  - Correct: Always add the $E^\circ$ values for reduction and oxidation of water to your comparison
- **Wrong:** Picking the most positive $E^\circ$ species for oxidation at the anode
  - Why it fails: Oxidation is the reverse of reduction, so the opposite rule applies
  - Correct: For oxidation at the anode, the species with the most negative (least positive) $E^\circ$ is oxidised preferentially
- **Wrong:** Assuming $E^\circ$ predictions work for all concentrations and temperatures
  - Why it fails: $E^\circ$ values are only valid under standard conditions
  - Correct: Check if the reaction is under standard conditions before making a prediction

## Cheatsheet

| Concept | Rule/Formula | Interpretation |
| --- | --- | --- |
| Calculate E°cell | $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ | Units: volts (V) |
| Spontaneous forward reaction | $E^\circ_{cell} > 0$ | ΔG° < 0, spontaneous under standard conditions |
| Non-spontaneous forward reaction | $E^\circ_{cell} < 0$ | ΔG° > 0, reverse reaction is spontaneous |
| Cathode (reduction) | Most positive E° | Reduced preferentially for similar concentrations |
| Anode (oxidation) | Most negative E° | Oxidised preferentially for similar concentrations |
| Common limitations | Non-standard conditions, activation energy, overpotential | Predictions may not match actual outcome |

## What's next

This subtopic connects electrochemistry to thermodynamics, and is a core foundation for all further electrochemistry content in CIE A-level Chemistry. Mastery of this content is essential for both multiple-choice and extended response questions, which frequently ask for spontaneity predictions, electrolysis product identification, and discussions of prediction limitations. Understanding how electrode potentials relate to reaction spontaneity also helps you connect concepts from redox, energetics, and electrochemistry across the syllabus. Next, you can explore how non-standard conditions affect cell potential, and learn about industrial electrochemical processes.

- [Further reaction kinetics](https://www.owlsprep.com/study/cie-9701-u20-overview/)
- [Rate equations and order of reaction](https://www.owlsprep.com/study/cie-9701-u20-rate-equations-and-order-of/)
- [Rate-determining step](https://www.owlsprep.com/study/cie-9701-u20-rate-determining-step/)

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