1 Faraday electricity was passed through Cu²⁺(1.5 M, 1 L) / Cu and 0.1 Faraday was passed through Ag⁺(0.2 M, 1 L) / Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is __________ mV (nearest integer)
Given : E° Cu²⁺ / Cu =0.34 V ; ; E° Ag⁺ / Ag =0.8 V ; ; (2 · 303 RT)/F =0.06 V

Solution

Chemistry ma’am · 4 steps

First let electrolysis eat into each solution, then put the new concentrations into Nernst.

1. Copper cell. Cu²⁺ + 2e⁻ → Cu needs 2 F per mole, so 1 F deposits 0.5 mol of Cu. In 1 L, [Cu²⁺] falls from 1.5 M to 1.0 M.
2. Silver cell. Ag⁺ + e⁻ → Ag needs 1 F per mole, so 0.1 F deposits 0.1 mol of Ag. In 1 L, [Ag⁺] falls from 0.2 M to 0.1 M.
3. Set up the cell. Silver has the higher reduction potential, so Ag is the cathode and Cu is the anode. The reaction is Cu + 2Ag⁺ → Cu²⁺ + 2Ag with n = 2, and E°_(cell) = 0.80 - 0.34 = 0.46 V.

   E_(cell) = 0.46 - 0.06/2 log[Cu²⁺]/[Ag⁺]² = 0.46 - 0.03 log 1/(0.1)² = 0.46 - 0.06

4. So. E_(cell) = 0.40 V, which is 400 mV.

The common slip: 413. Using the starting concentrations, 1.5 M and 0.2 M, ignores the metal deposited by electrolysis and gives about 413 mV.

The idea to keep. When electrolysis comes before the cell, convert Faradays to moles deposited and update every concentration first. Only then apply Nernst, with [Ag⁺] squared because n = 2.

Answer from the NTA final answer key. Solution by PrepWiser.