Question

A silver electrode is immersed in saturated$A{{g}_{2}}S{{O}_{4}}\left( aq \right)$. The potential difference between silver and the standard hydrogen electrode is found to be 0.711V.Determine ${{K}_{sp}}\left( A{{g}_{2}}S{{O}_{4}} \right)$. (Given$E_{{A{{g}^{+}}}/{Ag}\;}^{0}=0.799V$).

Answer 1

Hint:For solving this question we use the Nernst equation which is,
${{E}_{cell}}=E_{cell}^{0}-\dfrac{0.059}{n}\text{log}\dfrac{\left[ \text{Reactant} \right]}{\left[ \text{Product} \right]}$
The obtained ion concentration is used to calculate the solubility product value.

Complete step-by-step answer:In the question, it is given that a silver electrode is immersed in a saturated solution of silver sulfate and a cell reaction is happening in the cell. The standard reduction potential value of Ag cell and the potential difference between silver and standard hydrogen electrode is also given. We have to find the solubility product from the given data.
First, let us trace the reactions happening in cathode and anode cells.
Cathode: $A{{g}^{+}}+{{e}^{-}}\to Ag$
Anode: ${{H}_{2}}\to 2{{H}^{+}}+2{{e}^{-}}$
Multiply the reaction in the cathode with 2 to balance the number of electrons involved in the entire reaction.
Hence the net reaction will be :
Cathode: $2A{{g}^{+}}+2{{e}^{-}}\to 2Ag$
Anode: ${{H}_{2}}\to 2{{H}^{+}}+2{{e}^{-}}$
Net reaction: $2Ag+{{H}_{2}}\to 2A{{g}^{+}}+2{{H}^{+}}$
Now let's consider the Nernst equation for finding the ionic concentration of Ag ion formed.
We know that the Nernst equation is:
${{E}_{cell}}=E_{cell}^{0}-\dfrac{0.059}{n}\text{log}\dfrac{\left[ \text{Reactant} \right]}{\left[ \text{Product} \right]}$
Here ${{K}_{sp}}={{\left( 3.2\times {{10}^{-2}} \right)}^{2}}\left( \dfrac{3.2\times {{10}^{-2}}}{2} \right)$he given value${{E}_{cell}}$ is 0.711V. And the standard reduction potential of Ag is given a value of 0.799V.
The value of n is 2 and the reactant side has the hydrogen ions and the standard electrode potential of the hydrogen electrode is 1.
Now substitute all these values in the Nernst equation to get the concentration of Ag ion.
$0.711=0.799-\dfrac{0.059}{2}\times \log \dfrac{1}{{{\left[ A{{g}^{+}} \right]}^{2}}}$
$\left[ A{{g}^{+}} \right]=3.2\times {{10}^{-2}}$
Now we have the concentration of iron, and now have to calculate the solubility product.
The dissociation reaction $A{{g}_{2}}S{{O}_{4}}$is as follows: $A{{g}_{2}}S{{O}_{4}}\to 2A{{g}^{+}}+SO_{4}^{2-}$
i.e.${{K}_{sp}}={{s}^{2}}+\dfrac{s}{n}$
${{K}_{sp}}={{\left( 3.2\times {{10}^{-2}} \right)}^{2}}\left( \dfrac{3.2\times {{10}^{-2}}}{2} \right)$
${{K}_{sp}}=1.6\times {{10}^{-5}}$

Note:While doing such problems we should always trace the reactions happening in cathode and anode.
And it is important to know that in an anode always oxidation takes place and in cathode only reduction takes.
It is better to remember using a line AN OX-RED CAT.