Practice question
Question
What is the emf of the cell Cr(s) | Cr³⁺(0.002 M) || Ag⁺(0.05 M) | Ag(s) at 298 K? (Given: E°Cr³⁺/Cr = -0.74 V , E°Ag⁺/Ag = 0.80 V )
Explanation
E°cell = 0.80 - (-0.74) = 1.54 V . Ecell = 1.54 - (0.059/3) log ([Cr³⁺]/[Ag⁺]³) = 1.54 - 0.01967 log (0.002/0.000125) = 1.54 - 0.025 = 1.515 V .
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