Question

The standard reduction potential of $Pb$ and $Zn$ electrodes are $-0.126$ and $-0.763$ volts respectively. The e.m.f. of the cell $Zn\left|Z{n}^{2+}(0.1M)\Vert P{b}^{2+}(1M)\right|Pb$ is

• A. 0.637 V
• B. < 0.637 V
• C. > 0.637 V
• D. 0.889

Answer 1

Answer: (C)

$E=E^{\circ }-\frac{0.059}{n}\log \frac{\left[Z{n}^{+2}\right]}{\left[P{b}^{+2}\right]}$
$E^{\circ }=E_c^{\circ }-E_a^{\circ }=-0.126-(-0.763)$
$\Rightarrow -0.126+0.763=0.637$
$\Rightarrow 0.637-\frac{0.059}{2}\log \frac{0.1}{1}$
$=0.637-\frac{0.059}{2}\log \left(10^{-1}\right)$
$=0.637-\frac{0.059}{2}(-1)\log 10$
$\Rightarrow 0.637+\frac{0.059}{2}\times 1=0.637+0.0295$
$=0.6665$