Solveeit Logo
Login

Question

Question: The standard half-cell reduction potential for \(A{{g}^{+}}|Ag\) is 0.7991 V at \(25{}^\circ \)C. Gi...

The standard half-cell reduction potential for Ag+∣AgA{{g}^{+}}|Ag is 0.7991 V at 25∘25{}^\circ C. Given that the experimental value of Ksp 1.56×10−101.56\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

Explanation

Solution

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

& \Delta \text{G= -nFE } \\\ & \Delta \text{G= -RT ln K} \\\ & \\\ \end{aligned}$$ **Complete step by step answer:** When E is emf of cell, F is amount of charge passed, n is no. of electrons involved in reaction, G is Gibb’s energy of reaction then $\Delta \text{G= -nFE }$ From measurement of E of the cell, we can obtain Gibb’s Energy and vice versa can be also done. From Ksp value, Gibb’s Energy can be calculated using following Formula: $\Delta G=-RT\operatorname{ln Ksp}$ As given in data, $$A{{g}^{+}}+{{e}^{-}}\to Ag$$, number of moles of electron is one, so n=1 $\begin{aligned} & \Delta \text{G= -nFE} \\\ & \text{ } \\\ \end{aligned}$= -1 (96485) (0.7991) = -77.1KJ $$AgCl\to Ag+C{{l}^{-}}$$ $$\begin{aligned} & \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{aligned}$$ = 55.95KJ So, Gibb’s Energy required to calculate Standard half-cell reduction potential for Ag|AgCl cell can be obtained by adding both value of $\Delta G$ $\Delta G$= -77.10 + 55.95 = -21.15KJ So, standard half-cell potential can be calculated as follows: $${{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.