Electrical work done in one second is equal to electrical potential multiplied by total charge passed. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the EMF of the cell is E and nF is the amount of charge passed and ∆rG is the Gibbs energy of the reaction, then
∆rG = – nFEcell
It may be remembered that Ecell is an intensive parameter but ∆rG is an extensive thermodynamic property and the value depends on n. From the measurement of Eocell , we can obtain an important thermodynamic quantity, ∆rGo , standard Gibbs energy of the reaction.
What is the EMF of the cell if Eocell is 1.10 V?
Answer & explanation
Correct answer: option 4
EMF of the cell,
Ecell = Eocell - \(\frac{0.0591}{2}\)log\(\frac{Zn^{2+}}{Cu^{2+}}\)
Ecell = 1.10 - \(\frac{0.0591}{2}\)log\(\frac{0.1}{0.2}\)
Ecell = 1.1 - \(\frac{0.0591}{2}\)log\(\frac{1}{2}\) = 1.1 + \(\frac{0.0591}{2}\)log\(\frac{2}{1}\)
Ecell = 1.1 + \(\frac{0.0591×0.3010}{2}\) = 1.11 V