Question

The standard half-cell reduction potential for $A{{g}^{+}}|Ag$ is 0.7991 V at $25{}^\circ $C. Given that the experimental value of Ksp $1.56\times {{10}^{-10}}$ for AgCl, calculate the standard half-cell reduction potential for Ag|AgCl electrode.
(A) 0.2192V
(B) -0.2192V
(C) -1.2192V
(D) 1.219V

Answer 1

Hint: Ag undergoes reduction by accepting an electron. Standard potential of a cell can be calculated using standard Gibbs energy. Gibb’s energy can be calculated using the given Ksp value. Gibb’s energy is given by formula
\[\begin{align}
  & \Delta \text{G= -nFE } \\
 & \Delta \text{G= -RT ln K} \\
 & \\
\end{align}\]

Complete step by step answer:
When E is emf of cell, F is amount of charge passed, n is no. of electrons involved in reaction, G is Gibb’s energy of reaction then
$\Delta \text{G= -nFE }$

From measurement of E of the cell, we can obtain Gibb’s Energy and vice versa can be also done.
From Ksp value, Gibb’s Energy can be calculated using following Formula:
$\Delta G=-RT\operatorname{ln Ksp}$

As given in data,
\[A{{g}^{+}}+{{e}^{-}}\to Ag\], number of moles of electron is one, so n=1
$\begin{align}
 & \Delta \text{G= -nFE} \\
 & \text{ } \\
\end{align}$= -1 (96485) (0.7991)
                     = -77.1KJ
\[AgCl\to Ag+C{{l}^{-}}\]
\[\begin{align}
  & \Delta \text{G= -RT ln Ksp} \\
 & \text{ = -8}\text{.314}\times \text{298 }\times \text{ln (1}\text{.56}\times \text{1}{{\text{0}}^{10}}) \\
 & \\
\end{align}\] = 55.95KJ

So, Gibb’s Energy required to calculate Standard half-cell reduction potential for Ag|AgCl cell can be obtained by adding both value of $\Delta G$
$\Delta G$= -77.10 + 55.95 = -21.15KJ

So, standard half-cell potential can be calculated as follows:
\[{{E}^{0}}=\dfrac{-\Delta G}{nF}\dfrac{-21.15}{(1\text{) }96.485} = +0.2192\]V
 So, standard half-cell reduction potential for Ag|AgCl electrode is 0.2192V.
So, the correct answer is “Option A”.

Note: Electrical potential multiplied by total charge passed is equal to electrical work done in one second. When maximum work is to be obtained from a galvanic cell, then charge has to be passed reversibly. The reversible work done by galvanic cells is equal to a decrease in its Gibb’s energy.