Question

Calculate E°_cell for the following reaction at 298 K:

2Al_(s) + 3Cu⁺²(0.01M) → 2Al⁺³(0.01M) + 3Cu_(s)

Given: E_cell = 1.98V

Answer 1

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 

       E_cell = `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