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प्रश्न
Calculate E°cell for the following reaction at 298 K:
2Al(s) + 3Cu+2(0.01M) → 2Al+3(0.01M) + 3Cu(s)
Given: Ecell = 1.98V
उत्तर
Al(S) | Al3+ (aq) (0.01M) || Cu2+ (aq) (0.01M) | Cu(s) LHE (Al(s) → Al3+(aq) +3e-) × 2 (Oxidation at anode)
RHE [Cu2+ (aq) + 2e- → Cu(s) ] × 3 (reduction at cathode)
∴ n = 6
Ecell = `E°_(cell) - 0.0591/n log ([Al^(3+)]^2)/([Cu^(2+)]^3)`
`E°_(cell) = E_(cell) +0.0591/n log ([Al ^(3+)]^2)/([Cu^(2+)]^3)`
`= 1.98 + 0.0591/6 log (0.01)^2/((0.01)^3)`
`= 1.98 + 0.0591/6 log 10^2`
`= 1.98 + 0.0591/6 2 log 10`
`= 1.98 + 0.0591/6 xx 2 [ ∵ log 10 = 1]`
= 1.98 + 0.0197
E°cell = 1.9997 V
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