Question

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

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

Answer 1

Answer: (C)

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