
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
485.1k+ views
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.
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Which one of the following is a true fish A Jellyfish class 12 biology CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Write the difference between solid liquid and gas class 12 chemistry CBSE
