Q. 1.05 g of lead ore containing impurity of Ag was dissolved in ${HNO}_{3}$ and the volume was made 350 mL. An Ag electrode was dipped in the solution. $Pt\left( H_{2} \right)|\ H^{+}(1\ M)||\ {Ag}^{+}|Ag$ The $E_{cell}$ is 0.503 V at 298 K. The percent of lead in the ore is $\left( E_{{Ag}^{+}|Ag}^{{^\circ}} = 0.80\ V \right)$
A) 0.033%
B) 0.050%
C) 0.066%
D) 0.13%
Explanation
(a)
Cell reaction $:\frac{1}{2}H_{2}(g) + {Ag}^{+}(x) \rightleftharpoons Ag(x) + H^{+}(1\ M)$
$E = E^{{^\circ}} = - \frac{0.0591}{n}\log\frac{\left\lbrack H^{+} \right\rbrack}{\left\lbrack {Ag}^{+} \right\rbrack}$
$\ 0.503 = 0.80 - \frac{0.0591}{1}\log\frac{1}{x}$
$\ x = 9.43 \times 10^{- 6}\ M$
Number of moles of ${Ag}^{+}$ in 350 mL
$= \frac{9.43 \times 10^{- 6} \times 350}{1000} = 3.3 \times 10^{- 6}$
Mass of $Ag = 3.3 \times 10^{- 6} \times 108 = 3.56 \times 10^{- 4}\ g$
% of Ag in the ore $= \frac{3.56\ \times 10^{- 4}}{1.05} \times 100$
$= 0.0339\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.