For a cell, Cu(s)|Cu²⁺(0.001 M) ∥ Ag⁺(0.01 M)| Ag(s)
the cell potential is found to be 0.43 V at 298 K. The magnitude of standard electrode potential for Cu²⁺ / Cu is _________ × 10⁻² V.
[Given : E_(Ag + /Ag)^Θ = 0.80 V and 2.303RT/F = 0.06 V]

Solution

Chemistry ma’am · 4 steps

Write the cell reaction first, then let Nernst take you back from 0.43 V to the standard potential.

1. Cell reaction. Copper is the anode and silver is the cathode, so the reaction is Cu + 2Ag⁺ → Cu²⁺ + 2Ag. Two electrons are transferred, so n = 2.

   E_(cell) = E°_(cell) - 0.06/2 log[Cu²⁺]/[Ag⁺]²

2. Find the log term. [Cu²⁺]/[Ag⁺]² = 10⁻³/(10⁻²)² = 10⁻³/10⁻⁴ = 10, so log 10 = 1. Then 0.43 = E°_(cell) - 0.03 × 1, which gives E°_(cell) = 0.46 V.
3. Split the cell. E°_(cell) = E°_(Ag⁺/Ag) - E°_(Cu²⁺/Cu), so E°_(Cu²⁺/Cu) = 0.80 - 0.46 = 0.34 V.
4. So. The standard electrode potential of Cu²⁺/Cu is 0.34 V = 34 × 10⁻² V, so the answer is 34.

The common slip: 40. Leaving [Ag⁺] unsquared makes the ratio 0.1 instead of 10, so the log term flips sign and the answer comes out as 40, not 34.

The idea to keep. In the Nernst equation, every concentration is raised to its coefficient in the balanced reaction. Balance the electrons first, and n and the powers follow from that.

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