Practice question
Question
What is the emf of the cell Sn(s) | Sn²⁺(0.002 M) || Pb²⁺(0.02 M) | Pb(s) at 298 K? (Given: E°Sn²⁺/Sn = -0.14 V , E°Pb²⁺/Pb = -0.13 V )
Explanation
E°cell = -0.13 - (-0.14) = 0.01 V . Ecell = 0.01 - (0.059/2) log (0.002/0.02) = 0.01 + 0.0295 = 0.0395 V .
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