The electrode potential E((Zn2+/Zn)) of a zinc electrode at 25°C with an aqueous solution of 0.1M ZnSO4 is (E°((Zn2+/Zn)) = -0.76 V. Assume (2.303RT/F) = 0.06 at 298 K)

Question

The electrode potential E((Zn2+/Zn)) of a zinc electrode at 25°C with an aqueous solution of 0.1M ZnSO4 is (E°((Zn2+/Zn)) = -0.76 V. Assume (2.303RT/F) = 0.06 at 298 K) (A) + 0.73 (B) - 0.79 (C) - 0.82 (D) - 0.70

Tardigrade Admin · Accepted Answer

For Zn+2 longrightarrow Zn Ezn+ / zn =Ezn+2 / zn°-(2.303 R T/n F) log ([zn]/[zn+2]) =-0.76-(0.06/2) log (1/0.1)=-0.76-0.03 Ezn+2 / zn =-0.79 V

Q. The electrode potential E $_{(\frac{Z n ^{2 +}}{Z n})}$ of a zinc electrode at 25 $^{\circ}$ C with an aqueous solution of 0.1M $Z n S O_{4}$ is
(E $^{\circ}$ $_{(\frac{Z n ^{2 +}}{Z n})}$ = -0.76 V. Assume $\frac{2.303 RT}{F} =$ 0.06 at 298 K)

+ 0.73

- 0.79

- 0.82

- 0.70

+ 0.79

For $Z n^{+ 2} ⟶ Z n$
$E_{z_{n}}^{+} / z_{n} = E_{z_{n}^{+ 2} / z_{n}}^{\circ} - \frac{2.303 RT}{n F} lo g \frac{[ z _{n} ]}{[ z _{n}^{+ 2} ]}$
$= - 0.76 - \frac{0.06}{2} lo g \frac{1}{0.1} = - 0.76 - 0.03$
$E_{z_{n}}^{+ 2} / z_{n} = - 0.79 V$

Q. The electrode potential E(ZnZn2+​)​ of a zinc electrode at 25∘C with an aqueous solution of 0.1M ZnSO4​ is (E∘(ZnZn2+​)​ = -0.76 V. Assume F2.303RT​= 0.06 at 298 K)