Question

Give $E_{M{n}^{7+}|M{n}^{2+}}^0=1.5Vand{E}_{M{n}^{4+}|M{n}^{2+}}^0=1.2V,then{E}_{M{n}^{7+}|M{n}^{4+}}^0$ is:

• A. 0.3 V
• B. 0.1 V
• C. 1.7 V
• D. 2.1

Answer 1

The correct option is C 1.7 V

Since the final reaction is a half cell, we can use $\Delta G^0=\Delta G_1^0+\Delta G_2^0$
$\Delta G^0=-nFE$
$\Delta G^0=-nF{E}^0=-n_{1}F{E}_1^0+n_{2}F{E}_2^0$
$n{E}_{M{n}^{7+}|M{n}^{4+}}^0=n_1{E}_{M{n}^{7+}|M{n}^{2+}}^0-n_2{E}_{M{n}^{4+}|M{n}^{2+}}^0$
(since the second cell reaction needs to be reversed and added to the first)
$3E=5\times 1.5-2\times 1.2E=1.7V$