The electrochemical cells \(Zn| Zn^{2+} || Cu^{2+}| Cu\) and \(Fe | Fe^{2+} ||Cu^{2+} |Cu\) are connected in series. What will be the net emf of the cell at \(25^oC\)?
Given:
\(E^o\) of \(Zn^{2+}|Zn = -0.76 V\)
\(E^o\) of \(Cu^{2+}|Cu = +0.34 V\)
\(E^o\) of \(Fe^{2+}|Fe = -0.41 V\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → +1.85 V
When cells are connected in series, their emfs add up.
Cell 1, Zn|\( \ce{Zn^{2+}} \)||\( \ce{Cu^{2+}} \)|Cu: E°cell = E°right − E°left = 0.34 − (−0.76) = 1.10 V
Cell 2, Fe|\( \ce{Fe^{2+}} \)||\( \ce{Cu^{2+}} \)|Cu: E°cell = 0.34 − (−0.41) = 0.75 V
Net emf = 1.10 + 0.75 = +1.85 V