Question

Given, $P{b}^{2+}/Pb=-0.126V$; $Z{n}^{2+}/Zn=-0.763V$ Find the emf. of the following cell $Zn\left|Z{n}^{2+}(0.1M)\right|\left|P{b}^{2+}(1M)\right|Pb$.

• A. - 0.637
• B. + 0.637
• C. > 0.637
• D. + 0.889

Answer 1

Answer: (C)

$E_{\text{cell }}=E_{P{b}^{2+}/Pb}^0-E_{Z{n}^{2+}/Zn}^{\circ }$
$=-0.126-(-0.763)$
$=+0.637V$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \frac{\left[Z{n}^{2+}\right]}{\left[P{b}^{2+}\right]}$
$=0.637-\frac{0.0591}{2}\log 0.1$
$=0.637+0.02955=0.667V$