Consider the cell at 25°C
Zn | Zn²⁺ (aq), (1M) || Fe³⁺ (aq), Fe²⁺ (aq) | Pt(s)
The fraction of total iron present as Fe³⁺ ion at the cell potential of 1.500 V is x × 10⁻². The value of x is ______________. (Nearest integer)
(Given : E_(Fe³⁺/Fe²⁺)⁰ = 0.77V, E_(Zn²⁺/Zn)⁰ = - 0.76V)

Solution

Chemistry ma’am · 4 steps

Write the full cell reaction first, then let Nernst give you the ratio of Fe²⁺ to Fe³⁺.

1. Cell reaction. Zn is the anode and the Fe³⁺/Fe²⁺ couple is the cathode, so the reaction is Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺ with n = 2. The standard cell potential is E° = 0.77 - (-0.76) = 1.53 V.

   E = E° - 0.059/2 log[Zn²⁺][Fe²⁺]²/[Fe³⁺]² = 1.53 - 0.059 log[Fe²⁺]/[Fe³⁺]

2. Find the ratio. Since [Zn²⁺] = 1 M, the square and the 1/2 cancel. Then 1.500 = 1.53 - 0.059 log[Fe²⁺]/[Fe³⁺] gives log[Fe²⁺]/[Fe³⁺] = 0.03/0.059 = 0.508, so [Fe²⁺]/[Fe³⁺] = 3.22.
3. Fraction of Fe3+. Total iron is [Fe³⁺] + [Fe²⁺], so the fraction is 1/(1 + 3.22) = 1/4.22 = 0.237 = 23.7 × 10⁻².
4. So. x = 23.7, which rounds to 24.

The common slip: 31. Reporting the ratio [Fe³⁺]/[Fe²⁺] = 0.31 instead of the fraction of total iron gives 31, not 24.

The idea to keep. A fraction of the total is not the same as a ratio. Get the ratio r from Nernst, then the fraction is 1/(1 + r).

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