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**Hint:**. The Nernst equation is used to solve such problems. The equation is as follows:

${{E}_{cell}} = E_{cell}^{0}-\dfrac{RT}{nF}\ln Q$ ${{E}_{cell}} = E_{cell}^{0}-\dfrac{RT}{nF}\ln Q$

Q is the reaction quotient

- The number of electrons involved in the oxidation and reduction reaction is 2.

**Complete step by step answer:**

- In the question it is given that an alloy is dissolved in dilute $HN{{O}_{3}}$ and the volume is 100ml.The potential values of the cell is given. We have to find the percentage of Ag in the alloy which will be equal to the percentage of Ag deposited in the cathode.

- Let’s write a cell reaction for more clarity.

At anode, oxidation happens and the reaction taking place is:

${{H}_{2}}\to 2{{H}^{+}}+2{{e}^{-}}$

At cathode, reduction takes place and the reaction is:

$2Ag+2{{e}^{-}}\to 2Ag$

Combining the two equation, we get,${{H}_{2\left( g \right)}}+2Ag_{\left( aq \right)}^{+}\to 2{{H}^{+}}+2A{{g}_{\left( s \right)}}$

n=2 (no. of electrons involved).

And the Q value is,

$Q=\dfrac{{{\left[ {{H}^{+}} \right]}^{2}}}{{{\left[ A{{g}^{+}} \right]}^{2}}\left[ {{P}_{{{H}_{2}}}} \right]}$, where partial pressure of Hydrogen, it is taken as one and the concentration of $\left[ {{H}^{+}} \right]$ ions is taken as 1.

So the equation becomes, $Q = \dfrac{1}{{{\left[ A{{g}^{+}} \right]}^{2}}}$

Consider the Nernst equation,

${{E}_{cell}}=E_{cell}^{0}-\dfrac{RT}{nF}\ln Q$ or ${{E}_{cell}}=E_{cell}^{0}-\dfrac{2.303RT}{nF}\log Q$

Take and write the data provided in the question,

${{E}_{cell}} = 0.62V,E_{cell}^{0} = 0.80V,\dfrac{RT}{F} = 0.06$

Substitute all the values in the Nernst Equation,

$0.62=0.80-\dfrac{0.06}{2}\log Q$,

$-0.18=-\dfrac{0.06}{2}\log \dfrac{1}{{{\left[ A{{g}^{+}} \right]}^{2}}}$

$-0.36=-0.06\log \dfrac{1}{{{\left[ A{{g}^{+}} \right]}^{2}}}$

$\log \dfrac{1}{{{\left[ A{{g}^{+}} \right]}^{2}}}=6$

$\log \dfrac{1}{{{\left[ A{{g}^{+}} \right]}^{2}}}=6=-\log {{\left[ A{{g}^{+}} \right]}^{2}}$

$-2\log \left[ A{{g}^{+}} \right]=6$

$\log \left[ A{{g}^{+}} \right]=3$

$\left[ A{{g}^{+}} \right]={{10}^{-3}} = 0.001M$

- We got the concentration of the Ag deposited, we have to convert it into moles.

Millimoles of $A{{g}^{+}}$ = molarity $\times $ volume in ml = 0.001$\times $100 = 0.1millimole

We convert millimoles to mole, $mole = \dfrac{0.1}{1000} = {{10}^{-4}}mol$

$Weight\,of\,Ag\,deposited = no.\,of\,moles\times molecular\,mass = {{10}^{-4}}\times 108 = 0.0108$

- And now we have to convert this weight into percentage, to get the percentage of Ag in the alloy.

The weight of alloy is given as 1.08g.

- Percentage of Ag will be the weight of silver obtained divided by the weight of alloy multiplied with 100.

So the, $Percentage\,of\,Ag = \dfrac{0.0108}{1.08}\times 100=1$

**So, the correct answer is “Option D”.**

**Note:**By rearranging the Nernst equation and tracing the reduction and oxidation reactions, many parameters can be asked like to calculate potential of half-cell, to find the number of electrons involved etc.

- Caution should be given while calculating with log values and while taking antilog.

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