Question

The equilibrium constant for the disproportionation reaction 2Cu⁺(aq) rightleftharpoons Cu(s) + Cu²+(aq) at 25^∘C (EᶜircCᵤ⁺/Cᵤ=0.52V, EᶜircCᵤ²+/Cᵤ⁺=0.16V) is:

• A. \(6\times10^4\)
• B. \(6\times10^6\)
• C. \(1.2\times10^6\)
• D. 1.2×10⁻6

Hint:

Large positive E^∘ implies very large equilibrium constant.

Answer 1

The Correct Option is B

Step 1: Cell emf for disproportionation: E^∘ = Eᶜirccₐthₒdₑ - Eᶜircₐnode =0.52-0.16=0.36V
Step 2: Relation between E^∘ and K: log K = (nE^∘)/(0.059) Here n=1.
Step 3: log K=(0.36)/(0.059)≈6.1 ⟹ K≈10⁶.1≈6×10⁶