Redox and Electrode Potentials (OCR A-Level Chemistry A): Revision Notes

Predictions from electrode potentials

Introduction

Standard electrode potentials provide a powerful tool for predicting whether redox reactions are thermodynamically feasible. By comparing the electrode potential values ($E^{\circ}$) of different redox systems, we can determine which species will act as oxidising agents, which will act as reducing agents, and whether a reaction between them is likely to occur under standard conditions.

However, these predictions have important limitations that you must understand for the exam. While electrode potentials tell us if a reaction is thermodynamically possible, they cannot tell us how fast the reaction will occur or whether it will proceed under non-standard conditions.

Understanding the electrochemical series

Standard electrode potentials are measured under specific conditions: solutions at 1 mol dm⁻³ concentration, temperature of 298 K, and pressure of 100 kPa. These values are arranged in the electrochemical series, which allows us to make predictions about redox behaviour.

Consider the three redox systems shown in the table above:

Redox system	Half-equation	$E^{\circ}$ / V
A	$\text{Cr}^{3+}(aq) + 3e^- \rightleftharpoons \text{Cr}(s)$	$-0.77$
B	$\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)$	$+0.34$
C	$\text{Ag}^+(aq) + e^- \rightleftharpoons \text{Ag}(s)$	$+0.80$

From this arrangement:

• The strongest reducing agent is $\text{Cr}(s)$ (top right, most negative system)
• The strongest oxidising agent is $\text{Ag}^+(aq)$ (bottom left, most positive system)

Predicting feasible redox reactions

A redox reaction will be thermodynamically feasible when an oxidising agent from one system reacts with a reducing agent from another system, provided that:

Comparing redox systems

Let's examine how to predict reactions between different systems:

Comparing system C with systems A and B:

System C has the most positive $E^{\circ}$ value ($+0.80$ V), so $\text{Ag}^+(aq)$ is a strong oxidising agent. This should react with reducing agents from systems with less positive electrode potentials.

Therefore, $\text{Ag}^+(aq)$ should react with:

• $\text{Cr}(s)$ from system A (with $E^{\circ} = -0.77$ V)
• $\text{Cu}(s)$ from system B (with $E^{\circ} = +0.34$ V)

Comparing system B with systems A and C:

System B has an $E^{\circ}$ value of $+0.34$ V, which is more positive than system A but less positive than system C.

Therefore, $\text{Cu}^{2+}(aq)$ (the oxidising agent from B) should react with:

• $\text{Cr}(s)$ from system A only

The reaction between $\text{Cu}^{2+}(aq)$ and $\text{Ag}(s)$ is not predicted because system B has a less positive $E^{\circ}$ value than system C.

Comparing system A with systems B and C:

System A has the most negative $E^{\circ}$ value, so $\text{Cr}^{3+}(aq)$ will not act as an oxidising agent with either system B or C. There are no redox systems with less positive electrode potentials for it to oxidise.

Writing overall equations for redox reactions

Once you have predicted that a reaction is feasible, you need to be able to write the balanced overall equation. This involves combining the two relevant half-equations.

Method for combining half-equations:

1. Write the reduction half-equation exactly as shown (this is the system with the more positive $E^{\circ}$ value)
2. Reverse the oxidation half-equation (this is the system with the more negative $E^{\circ}$ value)
3. Balance the electrons by multiplying each half-equation by appropriate factors
4. Add the two half-equations together, canceling out the electrons

The same principle applies to any pair of redox systems:

• The system with the more positive (less negative) $E^{\circ}$ value reacts from left to right and gains electrons
• The system with the less positive (more negative) $E^{\circ}$ value reacts from right to left and loses electrons

Limitations of predictions using standard electrode potentials

While standard electrode potentials are useful for predicting feasibility, they have important limitations that you must understand. Predictions based on $E^{\circ}$ values can fail under certain circumstances.

Limitation 1: Reaction rate (kinetics)

This limitation mirrors a similar issue with using Gibbs free energy ($\Delta G$) to predict feasibility. Both $E^{\circ}$ and $\Delta G$ are thermodynamic quantities that tell us about the energy changes in reactions, but neither gives kinetic information about reaction rates.

Limitation 2: Concentration effects

Standard electrode potentials are measured using solutions with concentrations of 1 mol dm⁻³. Many reactions in practice involve either concentrated or dilute solutions with concentrations that differ significantly from this standard value.

When the concentration changes, the position of equilibrium in the redox system shifts according to Le Chatelier's principle, which affects the electrode potential value.

Any change to the electrode potential of individual systems will affect the overall cell potential and therefore the predicted feasibility of reactions.

Limitation 3: Other non-standard factors

Several other factors can cause actual electrode potentials to differ from standard values:

Relationship between electrode potentials and free energy

Standard electrode potentials are mathematically related to Gibbs free energy change ($\Delta G^{\circ}$), providing a connection between electrochemistry and thermodynamics.

Understanding the relationship

When $E^{\circ}_{\text{cell}}$ is positive:

• $\Delta G^{\circ}$ will be negative (because of the negative sign in the equation)
• A negative $\Delta G^{\circ}$ indicates a thermodynamically feasible (spontaneous) reaction

This confirms that for a feasible redox reaction, the oxidising agent must come from a system with a more positive electrode potential.

The smaller the value of $E^{\circ}_{\text{cell}}$ (i.e., the less positive), the less negative the value of $\Delta G^{\circ}$, indicating the reaction is less favorable thermodynamically.

Remember!