Question

You have a concentration cell in which the cathode hasa silver electrode with 0.10 M Ag⁺. The anode also hasa silver electrode with Ag⁺ (aq), 0.050 M, and 1.0 X 10^-3 M$\mathbf{Ag}{\mathbf{(}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}}^{\mathbf{3}\mathbf{-}}$. You read the voltage to be 0.76 V.

a. Calculate the concentration of Ag⁺ at the anode.
b. Determine the value of the equilibrium constant for the formation of$\mathbf{Ag}{\mathbf{(}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}}^{\mathbf{3}\mathbf{-}}$.

${\mathbf{Ag}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{S}}_{\mathbf{2}}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{Ag}{\mathbf{(}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}}^{\mathbf{3}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{K}\mathbf{=}\mathbf{?}$

Short answer

(a) The concentration of Ag⁺ at the anode is$1.4\times {10}^{-14} \mathrm{M}$

(b) The equilibrium constant for the formation of$\mathrm{Ag}{({\mathrm{S}}_2{\mathrm{O}}_3{)}_2}^{3-}$ is .$2.89\times {10}^{13}$

Step-by-step solution

Finding the concentration of Ag+

The electrochemical cell using the Nernst equation can be denoted as,

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}& =& {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{anode}}}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{cathode}}}\\ \log {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{anode}}& =& \frac{\mathrm{m}({{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}-{\mathrm{E}}_{\mathrm{cell}})}{0.0591}+\log {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{cathode}}\\ \log {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{anode}}& =& \frac{1(0-0.76)}{0.0591}-1\\ \log {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{anode}}& =& -13.86\end{array}$

Then,

${\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{anode}}=1.4\times {10}^{-14} \mathrm{M}$

Finding the equilibrium constant

The expression for equilibrium constant can be denoted as,

$\begin{array}{rcl}\mathrm{K}& =& \frac{\left[\mathrm{Ag}{{\left({\mathrm{S}}_2{\mathrm{O}}_3\right)}_2}^{3-}\right]}{\left[{\mathrm{Ag}}^{+}\right]{\left[{\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right]}^2}\\ \mathrm{K}& =& \frac{1\times {10}^{-3}}{(1.4\times {10}^{-14}){(0.05)}^2}\\ \mathrm{K}& =& 2.89\times {10}^{13}\end{array}$