Question

$E^\circ _{Cu} = 0.34\, V, E^\circ _{Zn }= - 0.76 \,V.A$ Daniell cell contains $0.1\,M \,ZnSO_4$ solution and $0.01\,M$ $CuSO_4$ solution at its electrodes. E.M.F. of the cell is

• A. 1.10 V
• B. 1.04 V
• C. 1.16 V
• D. 1.07 V

Answer 1

Answer: (D)

$Zn |Zn^{2+}||Cu^{2+}|Cu$
$Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$
$E^0_{cell} = E^0_{{Cu^{2+}/Cu}} - E^0 _{Zn^{2+}/Zn}$
$E_{cell} = E^0_{cell} - \frac{0.0591}{2}\,log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
$ = 1.10\,V -\frac {0.0591\,V}{2} \,log \,\frac{0.1}{0.01}$
$ = 1.10\,V - 0.05955\,V$
$ = 1.07045\,V$