The logarithm of equilibrium constant for the reaction Pd²⁺+4 Cl⁻ ⇌ PdCl₄²⁻ is ___________ (Nearest integer)
Given : (2.303 R T)/F=0.06 V
Pd_((aq))²⁺+2 e⁻ ⇌ Pd(s) E⊖=0.83 V
PdCl₄²⁻(aq)+2 e⁻ ⇌ Pd(s)+4 Cl⁻(aq) E⊖=0.65 V

Solution

Chemistry ma’am · 3 steps

Add the two half-reactions so that solid Pd cancels, then turn E°_(cell) into log K.

1. Build the reaction. Keep Pd²⁺ + 2e⁻ → Pd as the reduction. Reverse the second half-reaction to get the oxidation Pd + 4Cl⁻ → PdCl₄²⁻ + 2e⁻. Adding the two cancels both Pd and the electrons and leaves exactly Pd²⁺ + 4Cl⁻ ⇌ PdCl₄²⁻, with n = 2.
2. Cell potential. Both half-reactions move 2 electrons, so the standard potentials subtract directly: E°_(cell) = 0.83 - 0.65 = 0.18 V.

   log K = nE°_(cell)/0.06 = (2 × 0.18)/0.06

3. So. log K = 0.36/0.06 = 6, so the answer is 6.

The common slip: 3. Leaving out n = 2 gives 0.18/0.06 = 3, not 6. This reaction transfers two electrons.

The idea to keep. At equilibrium E_(cell) = 0, so the Nernst equation becomes log K = nE°/0.06 at 298 K. Always count the electrons transferred in the overall reaction.

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