Consider the cell
Pt(s)|H₂(g) (1 atm)|H + (aq,[H + ] = 1)||Fe³⁺(aq),Fe²⁺(aq)|Pt(s)
Given E_(Fe³⁺/Fe²⁺)ᵒ = 0.771 V and E_(H + /1/2 H₂)ᵒ = 0 V, T = 298 K
If the potential of the cell is 0.712 V, the ratio of concentration of Fe²⁺ to Fe³⁺ is _____________ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

The hydrogen half is fully standard, so the whole drop from 0.771 V comes from the iron ratio.

1. Cell reaction. The anode is the hydrogen electrode and the cathode is Fe³⁺/Fe²⁺. The overall reaction is H₂ + 2Fe³⁺ → 2H⁺ + 2Fe²⁺ with n = 2, and E°_(cell) = 0.771 - 0 = 0.771 V.
2. Write Nernst. Here [H⁺] = 1 and p_(H₂) = 1 atm, so only the iron terms stay in the quotient. Both iron terms are squared, and that square cancels the 2 in 0.059/2.

   E_(cell) = 0.771 - 0.059/2 log[Fe²⁺]²/[Fe³⁺]² = 0.771 - 0.059 log[Fe²⁺]/[Fe³⁺]

3. Put in numbers. 0.712 = 0.771 - 0.059 log[Fe²⁺]/[Fe³⁺], so log[Fe²⁺]/[Fe³⁺] = 0.059/0.059 = 1.
4. So. [Fe²⁺]/[Fe³⁺] = 10¹, which is 10.

The common slip: 100. Using n = 2 but not squaring the iron concentrations gives log ratio = 2, so the ratio comes out as 100, not 10.

The idea to keep. When one electrode is standard, only the other one's ratio moves the cell potential. Each factor of 10 in that ratio shifts E by 0.059/n V, with n counted per ion.

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