Question:

The electrode potential E$_{\left(\frac{Zn^{2+}}{Zn}\right)}$ of a zinc electrode at 25$^{\circ}$C with an aqueous solution of 0.1M $ZnSO_4$ is (E$^{\circ}$$_{\left(\frac{Zn^{2+}}{Zn}\right)}$ = -0.76 V. Assume $\frac{2.303RT}{F} =$ 0.06 at 298 K)

Updated On: Jul 29, 2022

0.73
-0.79
-0.82
-0.7

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The Correct Option is B

Solution and Explanation

For $Zn^{+2} \longrightarrow Zn$ $E_{z_{n}}^{+} / z_{n} =E_{z_{n}^{+2} / z_{n}}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[z_{n}\right]}{\left[z_{n}^{+2}\right]}$ $=-0.76-\frac{0.06}{2} \log \frac{1}{0.1}=-0.76-0.03$ $E_{z_{n}}^{+2} / z_{n} =-0.79\, V$

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Concepts Used:

Nernst Equation

This equation relates the equilibrium cell potential (also called the Nernst potential) to its concentration gradient across a membrane. If there is a concentration gradient for the ion across the membrane, an electric potential will form, and if selective ion channels exist the ion can cross the membrane.