Question

The potential of the cell for the reaction $M(s)+2H^{+}(1M)\to H_2(g)(1atm)+M^{2+}(0.1M)$ is 1 .500 V. The standard reduction potential for $M^{2+}/M$ couple is

• A. 0.1470 V
• B. 1.470V
• C. -1.47 V
• D. none of these

Answer 1

Answer: (C)

The cell is
$M\left|M^{2+}\left(0.1M\right)\right|\left|H^{+}\left(M\right)\right|H_2\left(1atm\right)$
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log[M^{+}]$
$1.5=E_{cell}^{\circ }-\frac{0.0591}{2}log(0.1)$
$1.5=E_{cell}^{\circ }+0.02955$
or $E_{cell}^{\circ }=1.50-0.02945V$
But $E_{cell}^{\circ }=E_{2H^{+}/H_2}^{\circ }-E_{M^{2+}/M}^{\circ }$
$\therefore 1.47=0-E_{M^{2+}/M}^{\circ }$
or $E_{M^{2+}/M}^{\circ }=-1.47V$