In a cell, the following reactions take place
Fe²⁺ → Fe³⁺ + e -, E_(Fe³⁺/Fe²⁺)ᵒ = 0.77 V; 2I - → I₂ + 2e -, E_(I₂/I -)ᵒ = 0.54 V
The standard electrode potential for the spontaneous reaction in the cell is x × 10⁻² V 298 K. The value of x is ____________. (Nearest Integer)

Solution

Chemistry ma’am · 3 steps

Let the higher reduction potential get reduced; then cathode minus anode gives the cell potential.

1. Pick the cathode. Both given values are reduction potentials. Fe³⁺/Fe²⁺ has the higher value (0.77 V), so Fe³⁺ is reduced at the cathode. I⁻ is oxidised to I₂ at the anode.
2. Spontaneous reaction. The reaction that actually runs is 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂. The iron half-equation in the question is written as oxidation, but in the spontaneous cell it runs the other way.

   E°_(cell) = E°_(cathode) - E°_(anode) = 0.77 - 0.54 = 0.23 V

3. So. E°_(cell) = 23 × 10⁻² V, so x = 23. Doubling the iron half-equation to balance electrons does not change E°, because it is an intensive property.

The common slip: -23. Taking the iodine couple as the cathode gives 0.54 - 0.77 = -0.23 V, that is -23. That is the non-spontaneous direction.

The idea to keep. For a spontaneous cell, the couple with the higher standard reduction potential is the cathode. Then E°_(cell) = E°_(cathode) - E°_(anode), and it must come out positive.

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