Question

The silver electrode is immersed in saturated ${\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4}$(aq). The potential difference between silver and the standard hydrogen electrode is found to be 0.711 V. Determine ${{\text{K}}_{{\text{Sp}}}}$ ${\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4}$. [Given ${\text{E}}_{{\text{A}}{{\text{g}}^ + }/{\text{Ag}}}^ \circ = 0.799{\text{V}}$].

Answer 1

Hint:Look at the values given in the questions, these two values[i.e., ${\text{E}}_{{\text{A}}{{\text{g}}^ + }/{\text{Ag}}}^ \circ $ and potential difference(or ${{\text{E}}_{{\text{cell}}}}$) ] belong to one equation, and by using that equation we can determine the concentration, which is what we need to find out ${{\text{K}}_{{\text{Sp}}}}$.

Formula used:
1. ${{\text{E}}_{{\text{cell}}}} = {\text{E}}_{{\text{cell}}}^ \circ - \dfrac{{0.059}}{{\text{n}}}{\text{log}}\dfrac{1}{{\left[ {{{\text{M}}^{{\text{n}} + }}} \right]}}$, where n = numbers of electrons,
[${{\text{M}}^{{\text{n + }}}}$] = concentration of reduction ion
${{\text{E}}_{{\text{cell}}}}$= potential difference in the cell (given in question)

2. ${\text{E}}_{{\text{cell}}}^ \circ = {\text{E}}_{{\text{reduction}}}^ \circ - {\text{E}}_{{\text{oxidation}}}^ \circ $ (given in question)

Complete answer:
Let's start with writing all the chemical reactions involved in this question.
First, we will write the chemical reaction for aqueous silver sulfate ${\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_{4{\text{(aq)}}}}$;
${\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4} \to 2{\text{A}}{{\text{g}}^ + } + {\text{SO}}_4^{2 - }$
Now Let’s look at the ${{\text{K}}_{{\text{Sp}}}}$ (equilibrium solubility constant) equation for ${\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4}$, the ${{\text{K}}_{{\text{Sp}}}}$ is written as concentration to products ${\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}\left[ {{\text{SO}}_4^{2 - }} \right]$divided by the concentration of reactants $\left[ {{\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4}} \right]$
$ \Rightarrow $${\text{Ksp = }}{\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}\left[ {{\text{SO}}_4^{2 - }} \right]$ [the stoichiometric coefficients are raised to power in the equation, and we take the concentration of solids and liquid as unity (1). Therefore, $\left[ {{\text{A}}{{\text{g}}_2}{\text{S}}{{\text{O}}_4}} \right]$= 1 ]
We will write the half cell reactions( i.e., the reaction occurring at the cathode and anode )
At cathode: The reduction half(acceptance of electrons) of the reaction takes place. Hence we write;
${\text{A}}{{\text{g}}^ + } + e \to {\text{Ag}}$
At anode: The oxidation half(giving away of electrons)of the reaction takes place. Hence we write;
${{\text{H}}_2} \to 2{{\text{H}}^ + } + 2e^-$
If we look at the two half cell reactions the number of electrons released is not equal to the number of electrons accepted. Therefore we have to balance out the number of the electron by multiplying 2 with the reduction half. Hence the reaction changes to;
$ \Rightarrow $ (${\text{A}}{{\text{g}}^ + } + e \to {\text{Ag}}$) x 2${\text{ = 2A}}{{\text{g}}^ + } + 2e \to 2{\text{Ag}}$
Now if we write the complete reaction of the cell, the electrons will cancel each other out, therefore the reaction can be written as;
${{\text{H}}_2} + 2{\text{A}}{{\text{g}}^ + } \to {\text{Ag + 2}}{{\text{H}}^ + }$
Here, n = 2 (since two electrons are involved)
$ \Rightarrow $${\text{E}}_{{\text{cell}}}^ \circ $= ${\text{E}}_{{\text{reduction}}}^ \circ - {\text{E}}_{{\text{oxidation}}}^ \circ $
$ \Rightarrow $${\text{E}}_{{\text{cell}}}^ \circ $ = ${\text{E}}_{{\text{A}}{{\text{g}}^ + }/{\text{Ag}}}^ \circ $- ${\text{E}}_{{\text{2}}{{\text{H}}^ + }/{{\text{H}}_2}}^ \circ $, the value of oxidation part i.e., ${\text{E}}_{{\text{2}}{{\text{H}}^ + }/{{\text{H}}_2}}^ \circ $ = 0
$ \Rightarrow $${\text{E}}_{{\text{cell}}}^ \circ $ = 0.799 – 0
$ \Rightarrow $${\text{E}}_{{\text{cell}}}^ \circ $ = 0.799
Now, we can calculate the concentration of ${\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}$ using the mentioned formula.
$ \Rightarrow $${{\text{E}}_{{\text{cell}}}} = {\text{E}}_{{\text{cell}}}^ \circ - \dfrac{{0.059}}{{\text{n}}}{\text{log}}\dfrac{1}{{{{\left[ {{\text{A}}{{\text{g}}^ + }} \right]}^2}}}$
$ \Rightarrow $ 0.711 = 0.799 – $\dfrac{{0.059}}{2}{\text{log}}\dfrac{1}{{{{\left[ {{\text{A}}{{\text{g}}^ + }} \right]}^2}}}$
$ \Rightarrow $ 0.711 – 0.799 = 0.00295${\text{log}}\dfrac{1}{{{{\left[ {{\text{A}}{{\text{g}}^ + }} \right]}^2}}}$
$ \Rightarrow $ – 0.0088 = – 0.00295${\text{log}}\dfrac{1}{{{{\left[ {{\text{A}}{{\text{g}}^ + }} \right]}^2}}}$, the two negative sigh will cancel each other
$ \Rightarrow $ $\dfrac{{0.0088}}{{0.00295}}$ = ${\text{log}}\dfrac{1}{{{{\left[ {{\text{A}}{{\text{g}}^ + }} \right]}^2}}}$
$ \Rightarrow $ 2.983 = log 1 – log ${\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}$, [ by using the operation ${\text{log}}\left( {\dfrac{{\text{x}}}{{\text{y}}}} \right) = {\text{log x - log y}}$]
$ \Rightarrow $ 2.983 = – log${\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}$we can convert 2.983 to 3 by rounding it off, to ease the calculation
$ \Rightarrow $ log ${\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}$= – 3
$ \Rightarrow $ ${\left[ {{\text{A}}{{\text{g}}^ + }} \right]^2}$=${10^{ - 3}}$, ( here we will use the log operation $ {\text{lo}}{{\text{g}}_{10}}{\text{x = y, x = 1}}{{\text{0}}^{\text{y}}}{\text{ }}$)
$ \Rightarrow \left[ {{\text{A}}{{\text{g}}^ + }} \right]$= $\sqrt {{{10}^{ - 3}}} $
$ \Rightarrow \left[ {{\text{A}}{{\text{g}}^ + }} \right]$= 3.1$ \times {10^{ - 2}}$
Since, $\left[ {{\text{A}}{{\text{g}}^ + }} \right]$= 3.1$ \times {10^{ - 2}}$, then ${\text{SO}}_4^{2 - }$= $\dfrac{{3.1 \times {{10}^{ - 2}}}}{2}$ = 1.55$ \times {10^{ - 2}}$, therefore value of ${{\text{K}}_{{\text{Sp}}}}$,
$ \Rightarrow $${{\text{K}}_{{\text{Sp}}}}{\text{ = }}{\left[ {3.1 \times {{10}^{ - 2}}} \right]^2}\left[ {1.55 \times {{10}^{ - 2}}} \right]$
$ \Rightarrow $${{\text{K}}_{{\text{Sp}}}}{\text{ = }}{\left[ {9.61 \times {{10}^{ - 2}}} \right]^2}\left[ {1.55 \times {{10}^{ - 2}}} \right]$
$ \Rightarrow $${{\text{K}}_{{\text{Sp}}}}$= 14.89 $ \times {10^{ - 6}}$
$ \Rightarrow $${{\text{K}}_{{\text{Sp}}}}$= 1.49 $ \times {10^{ - 5}}$

Hence the required value ${\text{Ksp}}$ is 1.49 $ \times {10^{ - 5}}$

Note:
To avoid confusion while writing the half cell reactions remember the term ‘red cat’(i.e., reduction at the cathode) and if the reduction is occurring at the cathode then oxidation will happen at the anode.