Question

A cell, Ag|$A{g^ + }$||$C{u^{2 + }}$|Cu, initially contains $1 M$ $A{g^ + }$and $1 M$ $C{u^{2 + }}$ions. The change in the cell potential after the passage of $9.65 A$ of current for 1 hour will be :
(A) 0.010 V
(B) 0.020 V
(C) 0.030 V
(D) None of these

Answer 1

Hint: We need to calculate the two values for cell potential. One will be calculated initially where the cell reaction is not taking place as given in the question. And the second cell potential is calculated after 1 hour when a large amount of current is supplied and the reaction is taking place. By comparing these two values of cell potential, the change can be easily calculated.

Complete step by step solution:
First, let us write the things given to us and the things we need to find out.
Given :
Initial concentration of $A{g^ + }$ $= 1M$
Initial concentration of $C{u^{2 + }}$ $= 1M$
Time period for which current passed = $1$ $hour$ = $3600$ $seconds$
Amount of current passed $= 9.65 A$
To find out :
Change in cell potential
We know that $E_{O{P_{Cu}}}^0$>$E_{O{P_{Ag}}}^0$
So, the cell in the reaction will not work.
But if we suppose the inverse cell of this, then it will work. The cell reaction for such a electrochemical cell can be written as -
$Cu \to C{u^{2 + }} + 2{e^ - }$
$2A{g^ + } + 2{e^ - } \to 2Ag$
The emf for this cell is -
${E_{cell}} = E_{O{P_{Cu}}}^0 + E_{R{P_{Ag}}}^0 + \dfrac{{0.059}}{2}\log \dfrac{{{{[A{g^ + }]}^2}}}{{[C{u^{2 + }}]}}$
We have
Initial concentration of $A{g^ + }$ $= 1M$
Initial concentration of $C{u^{2 + }}$$= 1M$
Thus, ${E_{cell}} = E_{O{P_{Cu}}}^0 + E_{R{P_{Ag}}}^0 + \dfrac{{0.059}}{2}\log 1M$
Log 1 = 0
So, ${E_{cell}} = E_{O{P_{Cu}}}^0 + E_{R{P_{Ag}}}^0$
This means, ${E_{cell}} = E_{cell}^0$
Let it be equation 1.
After the passing of current for 1 hour, during which the cell reactions are reversed. Thus, Ag metal will pass in solution while the $C{u^{2 + }}$ ions will be discharged.
The reactions can be written as-
$2Ag \to 2A{g^ + } + 2e$
$C{u^{2 + }} + 2{e^ - } \to Cu$
So, the amount of $A{g^ + }$will form = $\dfrac{{9.65 \times 3600}}{{96500}}$
the amount of $A{g^ + }$will form = $0.36 mole$
the amount of $C{u^{2 + }}$will discharge = $\dfrac{{9.65 \times 3600}}{{2 \times 96500}}$
the amount of $C{u^{2 + }}$will discharge = $0.18 mole$
So, the amount of $A{g^ + }$will be left $= 1 + 0.36$ $= 1.36 mole$
the amount of $C{u^{2 + }}$will be left $= 1 - 0.18$ $= 0.82 mole$
The emf can now be given as -
$\Rightarrow {E_{cell}} = E_{cell}^0 + \dfrac{{0.059}}{2}\log \dfrac{{{{(1.36)}^2}}}{{0.82}}$
On solving the equation, we get -
$\Rightarrow {E_{cell}} = E_{cell}^0 + 0.010{\text{ }}V$
Let it be equation 2.
Now comparing, equation 1 and equation 2; we have ${E_{cell}}$ value increased by $0.010 V$.

The option (A) is the correct answer.

Note: It must be noted that we know the oxidation potential of Copper is more than that of silver. So, the cell reaction given in the question is not possible initially when both have equal value of concentration. But after passing the current for 1 hour, the reaction is possible.