Question

What will be the $ E_{cell} $ for the given cell ?
$ Zn\left|Zn^{2+}\left(0.1\,M\right)\right|\left|Cu^{2+}\left(0.01\,M\right)\right|Cu $
Given : $ E^{\circ}_{Zn^{2+}Zn}=0.76\,V $ and $ E^{\circ}_{Cu^{2+}Cu}=0.34\,V $ .
Also predict whether the reaction is spontaneous or non-spontaneous.

• A. $ 1.07\, V $ and spontaneous
• B. $ -1.13\, V $ and non-spontaneous
• C. $ -1.07 \,V $ and non-spontaneous
• D. $ 1.13 \,V $ and spontaneous

Answer 1

Answer: (A)

$Zn\left|Zn^{2+}\left(0.1\,M\right)\right|\left|Cu^{2+}\left(0.01 M\right)\right|Cu$
$E_{\text{cell}}=E_{\text{cell}}^{\circ}-\frac{0.0591}{n} log \frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]}$
$E_{\text{cell}}^{\circ}=E^{\circ}_{\left(Cu^{2+}Cu\right)}-E^{\circ}_{\left(Zn^{2+}Zn\right)}$
$=0.34-\left(-0.76\right)=1.10\,V$
$E_{\text{cell}}=1.10-\frac{0.0591}{2} log \frac{0.1}{0.01}$
$E_{\text{cell}}=1.10-0.03=1.07$
As $E_{\text{cell}}$ is positive, the reaction is spontaneous