Consider the cell
Pt₍ₛ₎|H₂( g, 1 atm)| H⁺(aq, 1 M)|| Fe³⁺(aq), Fe²⁺(aq) | Pt(s)
When the potential of the cell is 0.712 V at 298 K, the ratio [Fe²⁺] /[Fe³⁺] is _____________. (Nearest integer)
Given : Fe³⁺+e⁻=Fe²⁺, E^θ Fe³⁺, Fe²⁺ | Pt=0.771
(2.303 RT)/F=0.06 V

Solution

Chemistry ma’am · 4 steps

The hydrogen side is all standard, so only the iron ratio is left in the Nernst equation.

1. Standard cell potential. The SHE on the left is the anode and has E° = 0. The Fe³⁺/Fe²⁺ electrode on the right is the cathode, so E°_(cell) = 0.771 - 0 = 0.771 V.
2. Write the reaction. The cell reaction is H₂ + 2Fe³⁺ → 2H⁺ + 2Fe²⁺ with n = 2. Here [H⁺] = 1 M and p_(H₂) = 1 atm, so only the iron terms stay in Q. Q holds the squared ratio and n = 2, so the two factors of 2 cancel.

   E_(cell) = 0.771 - 0.06/2 log([Fe²⁺]/[Fe³⁺])² = 0.771 - 0.06 log[Fe²⁺]/[Fe³⁺]

3. Put in numbers. Call the ratio r. Then 0.712 = 0.771 - 0.06 log r, so log r = 0.059/0.06 = 0.983. That gives r = 10^(0.983) ≈ 9.6.
4. So. The ratio [Fe²⁺]/[Fe³⁺] is about 9.6, which rounds to 10.

The common slip: 93. Using n = 2 but not squaring the iron ratio in Q gives log r = 1.97 and a ratio of about 93, not 10.

The idea to keep. When one half of the cell is a standard hydrogen electrode, its terms drop out of Q. The Nernst equation then depends only on the other electrode, with n set by that electrode's own electrons.

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