Question

What minimum decomposition potential is necessary to produce $C{l_2}$ gas in the reaction?
Given: $\left( {\dfrac{{2.303\,RT}}{F}\, = \,0.06} \right)$
$Sn{\,^{ + 2}}(IM)\, + \,2C{l^ - }(2M)\, \rightleftharpoons \,Sn\,(s)\, + \,C{l_2}\,(1\,atm)$
Given: $E_{S{n^{ + 2}},Sn}^ \circ = - 0.14\,V$ and $E_{C{l_2},\,C{l^{ - 1}}}^ \circ \, = \,1.4\,V$ .
A) $- 1.522\,V$
B) $- 1.558\,V$
C) $+ 1.522\,V$
D) $+ 1.358\,V$

Answer 1

Try to find out the $E_{cell}^ \circ$ in the question if the requirement of the examiner is to make this above reaction spontaneous then it will move forward and more amount of chlorine gas will be released. Now why we are solving this as finding $E_{cell}^ \circ$ because this reaction is in equilibrium and in equilibrium ${E_{cell}}$ is zero, thus we cannot solve the whole Nernst equation.

Complete answer:
We know that for a electrochemical cell reaction, there is relation between $E_{cell}^ \circ$ and ${E_{cell}}$ thus when we write the relation it will be like- ${E_{cell}} = \,E_{cell}^ \circ - \dfrac{{2.303RT}}{{nF}}\,\log \,\dfrac{{\left[ P \right]}}{{\left[ R \right]}}$
Here for the above reaction if we write the value of ${E_{cell}}$ as zero, it will becomes as $E_{cell}^ \circ = \,\dfrac{{2.303RT}}{{nF}}\,\log \,\dfrac{{\left[ P \right]}}{{\left[ R \right]}}$ now we have two ways for finding the minimum potential to make the reaction moving in forward direction after equilibrium. We can find out the $E_{cell}^ \circ$ for the reaction and the second way is to find out the value of terms on the right hand side.
To make the calculations easy for us, let’s try to find out the $E_{cell}^ \circ$ for reaction we know that it is equals to the difference between the $E_{cathode}^ \circ$ and $E_{anode}^ \circ$ .
So we have the formula for it, $E_{cell}^ \circ = \,E_{cathode}^ \circ - E_{anode}^ \circ$ as in every reaction a question arises that which part is acting like an anode and which is cathode.
Keep in mind that an electrode whose reduction potential is having higher value is termed as cathode while the electrode having lesser value of reduction value will be our anode. Here in the above question, among $E_{S{n^{ + 2}},Sn}^ \circ = - 0.14\,V$ and $E_{C{l_2},\,C{l^ - }}^ \circ \, = \,1.4\,V$ it is visible that in both reduction is shown and it is having greater value for chlorine thus on putting appropriate value in the above equation,
$E_{cell}^ \circ = \,E_{C{l_2},\,C{l^ - }}^ \circ - E_{S{n^{ + 2}},Sn}^ \circ$
$E_{cell}^ \circ = \,1.4\,V - ( - 0.14V)$
$E_{cell}^ \circ = \,1.54\,V$
This is the minimum potential for the production of chlorine gas therefore, this value is closest to option C. hence we get our answer.
Hence,Option C is the correct answer.

Note: As we know that in equilibrium the $\Delta G$ of reaction is zero, it means in the state of equilibrium the whole of Gibbs energy remains constant thus change is equal to zero. Therefore, the ${E_{cell}}$ also becomes zero by the help of this equation, $\Delta G = \, - nF{E_{cell}}$ . If you put zero in place of ${E_{cell}}$ in Nernst equation, to make the reaction moving in forward reaction you have to find out the $E_{cell}^ \circ$ .