1. Tardigrade
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  3. Chemistry
  4. The potential of the cell for the reaction M(s)+2H+(1M) → H2(g) (1 atm) + M2+(0.1 M) is 1 .500 V. The standard reduction potential for M2+/M couple is

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

Electrochemistry

Solution:

The cell is
$M \left|M^{2+}\left(0.1M\right)\right| \left|H^{+}\left(M\right)\right|H_{2} \left(1 atm\right)$
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} log[M^{ +}]$
$ 1.5= E^{\circ}_{cell} - \frac{0.0591}{2} log (0.1)$
$ 1 . 5 = E^{\circ}_{cell} + 0.02955$
or $E^{\circ}_{cell} = 1.50 - 0.02945\,V$
But $E^{\circ}_{cell} = E^{\circ}_{2H^+ /H_2} - E^{\circ}_{M^{2+}/M}$
$\therefore 1 .47 = 0 - E^{\circ}_{M^{2+}/M}$
or $E^{\circ}_{M^{2+}/M} = -1.47\,V$