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  • The equilibrium constant for the reaction, <img alt="2 \mathrm{Fe}^{3+}+\mathrm{H}_2(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}^{2+}+2 \mathrm{H}^{+}" src="https://entrancecorner.oncodecogs.com

The equilibrium constant for the reaction, 2 \mathrm{Fe}^{3+}+\mathrm{H}_2(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}^{2+}+2 \mathrm{H}^{+} at 25^{\circ} \mathrm{C} is 10^{22} . What is the value of the standard reduction potential of \mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+} half - cell at 25^{\circ} \mathrm{C} ?

Option: 1

0.02 V


Option: 2

0.2 V


Option: 3

0.03 V


Option: 4

0.3 V


Answers (1)

best_answer

We know that the equilibrium constant, K, can be related to the standard cell potential,E^{\circ }  , by the Nernst equation:

E=E^{\circ}-\frac{n F}{R T} \ln K

At 25^{\circ} \mathrm{C}, \mathrm{R}=8.314 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}, \mathrm{T}=298 \mathrm{~K}, n=2 and \mathrm{F}=96485 \mathrm{C} / \mathrm{mol}.

Substituting these values along with the given \mathrm{K=10^{22}}  into the equation, we get:

\ln K=(R T / 2 F)\left(E^{\circ}-E\right)

Solving for E^{\circ}, we get :

E^{\circ}=E+\left(\frac{R T}{2 F}\right) \ln K

Substituting the values, we get:

\begin{aligned} E^{\circ} & =0+\frac{(8.314 \times 298)}{(2 \times 96,485)} \times \ln \left(10^{22}\right) \\ \\E^{\circ} & =0.03 \mathrm{~V} \end{aligned}

Therefore, the value of the standard reduction potential of \mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+} half - cell at 25^{\circ} \mathrm{C} is 0.03 V , which is option (3).

Posted by

jitender.kumar

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