Q. The potential of the cell for the reaction, $\ M\ (s)\ + \ {2H}^{+}\ (1M) \longrightarrow \ H_{2}\ (g)\ (1\ atm)\ + \ M^{2 +}\ (0.1\ M)’$ is 1.500 V. The standard reduction potential for $M^{+}/\ M\ (s)$ couple is
A) 0.1470 V
B) 1.470 V
C) 14.70 V
D) None of these
Explanation
(b)
Given, product = 0.1 M and reactant = 1 M
$E_{cell}^{{^\circ}} = \ E_{cell}^{{^\circ}} - \ \frac{0.0591}{2}\log\frac{\lbrack products\rbrack}{\lbrack reactants\rbrack}$
$1.50\ = \ E_{cell}^{{^\circ}}\ - \ 0.02955\ \log\ \left\lbrack \frac{0.1}{1} \right\rbrack$
$E_{cell}^{{^\circ}} = \ 1.470\ V$
$E_{cell}^{{^\circ}} = \ E_{H^{+}\ /H_{2}}^{{^\circ}}\ - \ E_{M\ /M^{2 +}}^{{^\circ}}$
${E{^\circ}}_{M\ /M^{2 +}}\ = \ - \ 1.470$
So,$\ E_{M^{2 +}\ /M}^{{^\circ}}\ = \ 1.470\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.