Given the standard reduction potentials at 298 K : E^ _ Cu²⁺/Cu = 0.327 V E^ _ Cu²⁺/Cu^+ = 0.150 V Assume 2.303RT F = 0.059 V . The value of ₁₀ K_c for the disproportionation reaction 2Cu⁺(aq) Cu²⁺(aq) + Cu(s) is:

Step-by-step solution

First, we calculate the standard reduction potential for the Cu⁺/Cu half-cell. The half-reactions are: Cu²⁺ + 2e⁻ Cu G^ ₁ = -2F(0.327) = -0.654F Cu²⁺ + e⁻ Cu⁺ G^ ₂ = -1F(0.150) = -0.150F Subtracting the second reaction from the first gives the required half-reaction: Cu⁺ + e⁻ Cu G^ ₃ = G^ ₁ - G^ ₂ = -0.654F - (-0.150F) = -0.504F Since G^ ₃ = -1FE^ _ Cu⁺/Cu , we have: E^ _ Cu⁺/Cu = 0.504 V Next, we set up the disproportionation reaction: 2Cu⁺ Cu²⁺ + Cu This can be split into: Oxidation (Anode): Cu⁺ Cu²⁺ + e⁻ (E^ _ a