Practice question
Question
What is the emf of the cell Pb(s) | Pb²⁺(0.005 M) || Ag⁺(0.05 M) | Ag(s) at 298 K? (Given: E°Pb²⁺/Pb = -0.13 V , E°Ag⁺/Ag = 0.80 V )
Explanation
E°cell = 0.80 - (-0.13) = 0.93 V . Ecell = 0.93 - (0.059/2) log (0.005/0.05²) = 0.93 - 0.0295 = 0.9005 V .
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