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  • What is the cell potential of a reaction involving a nickel electrode and a silver electrode at 298 K, where the concentration of nickel ion is 0.05 M and the concentration of silver ion is 0.01 M,

What is the cell potential of a reaction involving a nickel electrode and a silver electrode at 298 K, where the concentration of nickel ion is 0.05 M and the concentration of silver ion is 0.01 M, according to the Nernst equation?

\begin{aligned} & E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.25 \mathrm{V\ and\ E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}= 0.80 \end{aligned}

Option: 1

1.05 V


Option: 2

0.75 V


Option: 3

0.55 V


Option: 4

0.30 V


Answers (1)

best_answer

The Nernst equation is given by:

E=E^{\circ}-\frac{R T}{n F} \ln (Q)

The balanced cell reaction is :

\mathrm{Ni}(s)+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(s)

The reaction quotient Q is given by:

Q=\frac{\left[\mathrm{Ni}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^2}

Substituting the given concentrations, we get:

\begin{aligned} &Q=\frac{0.05}{0.01^2}\\ \\&Q=500 \end{aligned}

The cell potential is given by:

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

Substituting the given values, we get:

\begin{aligned} & E=0.80-\frac{0.0257}{2} \ln (500)-(-0.25) \\ \\& E=0.55 \mathrm{~V} \end{aligned}

Therefore, the cell potential of the given reaction is 0.55 V. Option (3) is the correct answer

Posted by

vishal kumar

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