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For the reaction at 298 \mathrm{~K}, \mathrm{Ag}(s)   \mathrm{Pb}(s)\left|\mathrm{Pb}^{2+}(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(1.0 \mathrm{M})\right| the reduction potential of \mathrm{Pb}^{2+} / \mathrm{Pb} half-cell is -0.13 \mathrm{~V}, and the standard reduction potential of \mathrm{Ag}^{+} / \mathrm{Ag} half-cell is 0.80 \mathrm{~V}. The cell potential \left(E_{\text {cell }}\right) of the given cell is closest to:

Option: 1

\begin{aligned} & -0.93 \mathrm{~V} \\ \end{aligned}


Option: 2

-0.53 \mathrm{~V}


Option: 3

-0.67 \mathrm{~V}


Option: 4

-0.97 \mathrm{~V}


Answers (1)

best_answer

The cell potential \left(E_{\text {cell }}\right)can be calculated using the Nernst equation:

E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{R T}{n F} \ln Q

where E_{\text {cell }}^{\circ} is the standard cell potential, \mathrm{R} is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.
At 298 \mathrm{~K}, the reaction quotient Q can be calculated as:

Q=\frac{\left[\mathrm{Ag}^{+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=\frac{1.0 \mathrm{M}}{0.01 \mathrm{M}}=100

Substituting the given values into the Nernst equation and solving for E_{\text {cell }}^{\circ} , we get:

\begin{aligned} & E_{\text {cell }}=0.80 \mathrm{~V}-\frac{0.0592 \mathrm{~V}}{2} \log (100)- (-0.13 \mathrm{~V}) \\ \\& E_{\text {cell }}=-0.53 \mathrm{~V} \end{aligned}

Hence, the closest answer is (2)-0.53 \mathrm{~V}.

Posted by

sudhir.kumar

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