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Calculate E°cell for the following reaction at 298K: 2Al(s) + 3Cu+2(0.01M) → 2Al+3(0.01M) + 3Cu(s) - Chemistry

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Numerical

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

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Solution

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

cell  = 1.9997 V 

Concept: Galvanic Cells - Introduction
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