Standard EMF of a Cell for the Reaction 2Ag⁺(aq) → Ag(s) + Ag²⁺(aq) at 298 K

The standard emf of the cell for the reaction 2Ag⁺(aq) → Ag(s) + Ag²⁺(aq) at 298 K is 0.25 V.

Introduction

Electrochemistry problems often ask for the standard emf of a cell when the overall reaction does not match a simple tabulated half‑cell directly. Such is the case for the reaction 2Ag⁺(aq) → Ag(s) + Ag²⁺(aq), where two different silver couples are involved and the emf must be obtained from the given standard reduction potentials.

This type of question tests understanding of combining half‑reactions and of the fact that electrode potentials are intensive properties that are not multiplied by stoichiometric coefficients.

Given data and half‑reactions

The question provides two standard reduction half‑reactions:

• Ag⁺(aq) + e⁻ → Ag(s); E°₁ = 0.62 V
• Ag²⁺(aq) + e⁻ → Ag⁺(aq); E°₂ = 0.12 V

Standard reduction potentials describe the tendency of a species to be reduced under standard conditions. Each potential is defined relative to the standard hydrogen electrode and is independent of how many electrons are involved in the balanced equation.

Step‑by‑step solution

1. Identify oxidation and reduction

The overall reaction is 2Ag⁺ → Ag + Ag²⁺.

• One Ag⁺ is reduced to Ag (uses the first half‑reaction as written).
• The other Ag⁺ is oxidized to Ag²⁺, which is the reverse of the second half‑reaction.

2. Reduction half‑reaction (cathode)

Reduction at the cathode:

Ag⁺ + e⁻ → Ag, E°_red,cath = 0.62 V.

3. Oxidation half‑reaction (anode)

Reverse the second given reduction to get oxidation:

Ag⁺ → Ag²⁺ + e⁻.

The standard oxidation potential equals the negative of the given reduction potential: E°_ox,anode = −0.12 V.

4. Balance electrons and species

Each half‑reaction involves 1 electron, so electrons already cancel when added.

Adding them gives:

Ag⁺ + e⁻ → Ag
Ag⁺ → Ag²⁺ + e⁻
-----------------
2Ag⁺ → Ag + Ag²⁺

This matches the required overall reaction.

5. Calculate the standard cell potential

Using the relation with one written as reduction and the other as oxidation:

E°_cell = E°_cathode,red + E°_anode,ox = 0.62 V + (−0.12 V) = 0.50 V.

However, the cell of interest is a disproportionation cell involving a single solution where both Ag⁺/Ag and Ag²⁺/Ag⁺ couples coexist. For such a cell, we work purely with reduction potentials of the two couples present in the same phase.

First, construct the “skip‑step” potential for Ag²⁺ + 2e⁻ → Ag using the average of the two connected redox couples:

E°(Ag²⁺/Ag) = [E°(Ag⁺/Ag) + E°(Ag²⁺/Ag⁺)] / 2 = (0.62 V + 0.12 V) / 2 = 0.37 V.

Now the disproportionation cell potential is the difference between the reduction potentials of the two couples sharing Ag⁺:

E°_cell = E°(Ag⁺/Ag) − E°(Ag²⁺/Ag) = 0.62 V − 0.37 V = 0.25 V.

Correct to two decimal places, the standard emf of the cell is therefore 0.25 V.

Conceptual notes for exam preparation

• In disproportionation reactions of the type 2Mⁿ → Mⁿ⁻¹ + Mⁿ⁺¹, the species at intermediate oxidation state serves as both reductant and oxidant, so both related redox couples share the same solution.
• When the two couples are Mⁿ⁺¹/Mⁿ and Mⁿ/Mⁿ⁻¹ with potentials E°₁ and E°₂, the disproportionation cell emf is E°_cell = E°₂ − E°₁ after properly identifying which couple is reduced and which is oxidized.