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E°cell = 0.80 - (-0.25) = 1.05 V . Ecell = E°cell - (0.059/2) log ([Ni²⁺]/[Ag⁺]²) = 1.05 - (0.059/2) log (0.001/(0.002)²) = 0.989 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
E°cell = 0.80 - (-0.25) = 1.05 V . Ecell = E°cell - (0.059/2) log ([Ni²⁺]/[Ag⁺]²) = 1.05 - (0.059/2) log (0.001/(0.002)²) = 0.989 V .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
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