The electrochemical cells \(Zn|Zn^{2+}||Cu^{2+}|Cu\) are connected in series. What will be the net emf of the cell at \(27^oC\)?

Given:

\(E^o \text{ of }Zn^{2+}|Zn = - 0.76 V\)

\(E^o \text{ of }Cu^{2+}|Cu = + 0.34 V\)

\(E^o \text{ of }Fe^{2+}|Fe = - 0.41 V\)

+ 1.85 V

The correct answer is (1) + 1.85 V.

The given electrochemical cell is \(Zn|Zn^{2+}||Cu^{2+}|Cu\).

Given:

\(E^o \text{ of }Zn^{2+}|Zn = - 0.76 V\)

\(E^o \text{ of }Cu^{2+}|Cu = + 0.34 V\)

\(E^o \text{ of }Fe^{2+}|Fe = - 0.41 V\)

So,

\(E^o_{Cu^{2+}/Cu} – E^o_{Zn^{2+}/Zn} = +0.34 – (– 0.76) = 1.10 V\)

Also,

\(E^o_{Cu^{2+}/Cu} – E^o_{Fe^{2+}/Fe} = +0.34 – (– 0.41) = 0.75V\)

Since the two electrodes are connected in series, the net emf of cell \(= 1.10 + 0.75 = +1.85 V\)