Question

For the reaction.
$2AgCl\left( s \right) + {H_2}\left( g \right)\left( {1atm} \right) \to 2Ag\left( s \right) + 2{H^ + }\left( {0.1M} \right) + 2C{l^ - }\left( {0.1M} \right)$
$\Delta {G^o} = - 43600J$ at ${25^o}C$
Calculate the e.m.f of the cell.
$\left[ {\log {{10}^{ - n}} = - n} \right]$

Answer 1

Hint: Production of $Ag$ from the $AgCl$ molecules requires a redox reaction to take place. The calculation of e.m.f depends on the value of $E_{cell}^o$ that needs to be calculated at first and then from there the ${E_{cell}}$ can be calculated.

Complete step by step answer:
The given reaction of conversion is a redox process where the reduction and oxidation occur based on changes in the oxidation number.
Reduction process is $AgCl\left( s \right) \to Ag(s)$
Oxidation process is ${H_2}\left( g \right) \to 2{H^ + }$
These are the two processes taking place in the reaction and hence based on this process $E_{cell}^o$ can be calculated using the given calculated values. Here $F = 96500$ which is the Faraday’s constant and $n$ is the number of molecules that are involved in the reaction.
$E_{cell}^o = \dfrac{G}{{nF}}$
Putting the values in the equation we get,
$E_{cell}^o = \dfrac{{ - ( - 43600)}}{{2 \times 96500}}$
$ \Rightarrow E_{cell}^o = 0.23$
Therefore, now the $E_{cell}^o$ is calculated and from there the ${E_{cell}}$ can be easily calculated. Based on this calculation the e.m.f of the cell can be formulated. The products are taken as the numerator using the $2$ molecules of each as the power of each.
${E_{cell}} = E_{cell}^o - \dfrac{{0.0591}}{n}\log \dfrac{{{{\left[ {{H^ + }} \right]}^2}{{\left[ {C{l^ - }} \right]}^2}}}{1}$
Taking molarity values as concentration \[ \Rightarrow {E_{cell}} = 0.23 - \dfrac{{0.0591}}{2}\log {\left( {\dfrac{1}{{10}}} \right)^2}{\left( {\dfrac{1}{{10}}} \right)^2}\]
$ \Rightarrow {E_{cell}} = 0.23 - \dfrac{{0.0591}}{2}\log {\left( {{{10}^{ - 1}}} \right)^4}$
\[ \Rightarrow {E_{cell}} = 0.23 - \dfrac{{0.0591}}{2}\log \left( {{{10}^{ - 4}}} \right)\]
Since it is already given before that, $\left[ {\log {{10}^{ - n}} = - n} \right]$
Hence $ \Rightarrow {E_{cell}} = 0.23 - \dfrac{{0.0591}}{2} \times \left( { - 4} \right)$
$ \Rightarrow {E_{cell}} = 0.23 - \left\{ {0.0591 \times \left( { - 2} \right)} \right\}$
$ \Rightarrow {E_{cell}} = 0.23 + 0.1182$
$ \Rightarrow {E_{cell}} = 0.3482$

Therefore the e.m.f. of the cell for the given set of conditions is $0.3482$. This is the value of the emf of the specific cell under the condition where $AgCl$ is converted to $Ag$. Therefore, formulating the redox reaction is important in the process so that $E_{cell}^o$ can be calculated as the oxidation half cell and reduction half cell of the process can be easily formulated.

Note: Calculating the emf of a cell requires the calculation of $E_{cell}^o$ which includes the difference between the ${E^o}$ at the cathode and ${E^o}$ at the anode of the cell. This shows the e.m.f of the electrophoretic cell which makes the change from the $AgCl$ to $Ag$ during the process.