Cl2 (g) + 2e- ⇌ 2Cl- (aq) Eꝋ = +1.36 V
Cu2+ (aq) + 2e- ⇌ Cu (s) Eꝋ = +0.34 V
Cl2 (g) + 2e- → 2Cl- (aq)
Cu (s) → Cu2+ (aq) + 2e-
Cu (s) + Cl2 (g) → 2Cl- (aq) + Cu2+ (aq)
OR
Cu (s) + Cl2 (g) → CuCl2 (s)
A reaction is feasible when the standard cell potential Eꝋ is positive
You may have to apply your understanding (from the above worked example) to questions with more than 2 equations
Eθ ∝ ΔStotal
Eθ ∝ lnK
You are expected to be aware of the two directly proportional relationships described
However, you are not expected to work with the Faraday based Gibbs free energy equation
ΔG = –nF Eθcell
V3+ (aq) + e- ⇌ V2+ (aq) Eθ = +0.26 V
Cu2+ (aq) + 2e- ⇌ Cu (s)Eθ = 0.34 V
2H+ (aq) + 2e- ⇌ H2 (g)Eθ = 0.00 V
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