Consider the following redox reaction :
MnO₄⁻+H⁺+H₂ C₂ O₄ ⇌ Mn²⁺+H₂ O+CO₂
The standard reduction potentials are given as below (E_(red)⁰) :
E_(MnO₄⁻ / Mn²⁺)°=+1.51 V ; ; E_(CO₂ / H₂ C₂ O₄)°=-0.49 V
If the equilibrium constant of the above reaction is given as K_(eq)=10ˣ, then the value of x= __________ (nearest integer)

Solution

Chemistry ma’am · 3 steps

Get E°_(cell) and the total electrons moved, then log K falls straight out.

1. Cell potential. MnO₄⁻ is reduced (cathode) and H₂C₂O₄ is oxidised (anode). So E°_(cell) = E°_(cathode) - E°_(anode) = 1.51 - (-0.49) = 2.00 V.
2. Count electrons. Mn goes from +7 to +2, gaining 5 electrons. Each C goes from +3 to +4, so one H₂C₂O₄ loses 2 electrons. Balancing gives 2MnO₄⁻ + 6H⁺ + 5H₂C₂O₄ → 2Mn²⁺ + 8H₂O + 10CO₂, so n = 10.

   log K = (n E°_(cell))/0.0591 = (10 × 2.00)/0.0591 = 338.4

3. So. K_(eq) = 10^(338.4), so x to the nearest integer is 338.

The common slip: 169. Taking n = 5, the electrons gained by one MnO₄⁻, gives log K ≈ 169. The balanced equation has 2 MnO₄⁻, so 10 electrons are transferred.

The idea to keep. At equilibrium Δ G = 0, so log K = nE°/0.0591 at 298 K. Take n from the fully balanced equation: it is the total number of electrons transferred, not the count per ion.

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