Practice question
Question
The emf of a cell Fe(s) | Fe²⁺(0.001 M) || Cu²⁺(0.1 M) | Cu(s) at 298 K is (Given: E°Fe²⁺/Fe = -0.44 V , E°Cu²⁺/Cu = 0.34 V )?
Explanation
E°cell = 0.34 - (-0.44) = 0.78 V . Ecell = 0.78 - (0.059/2) log (0.001/0.1) = 0.78 + 0.059 = 0.839 V .
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