Question

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

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

Answer 1

Answer: (A)

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