Redox reaction Zn(s) + Cu2+ (0.1M) → Zn2+ (1M) + Cu(s) takes place in a cell with Eocell = 1.1V then what is Ecell for the cell at 298K?
Answer & explanation
Correct answer: option 3
Ecell = Eocell - \(\frac{0.059}{2}\)log\(\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
Ecell = 1.1 - \(\frac{0.059}{2}\)log\(\frac{[1]}{[0.1]}\)
Ecell = 1.1 - 0.0295 log10 [∴log10 = 1]
Ecell = 1.1 - 0.0295 = 1.0705 V