Question

A cell, $Ag|A{g}^{+}||C{u}^{2+}|Cu$, initially contains 1 M Ag${}^{+}$ and 1 M Cu${}^{2+}$ 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

The correct option is A 0.010 V

Note that the given cell will not work as electrochemical cells since,
$E_{O{P}_{Cu}}^o>E_{O{P}_{Ag}}^o$
The equation for electrochemical cells will be: (i)
$Cu\to C{u}^{2+}+2e$
$2A{g}^{+}+2e\to 2Ag$
Thus, e.m.f. of cell $Cu|C{u}^{2+}||A{g}^{+}|$ Ag will be :
$E_{cell}=E_{O{P}_{Cu}}^o+E_{R{P}_{Ag}}^o+\frac{0.059}{2}log\frac{[A{g}^{+}{]}^2}{\left[C{u}^{2+}\right]}$
$\therefore \left[A{g}^{+}\right]=1M$ and $\left[C{u}^{2+}\right]=1M$
$\therefore E_{cell}=E_{O{P}_{Cu}}^o+E_{R{P}_{Ag}}^o$ $(E_{cell}^o=E_{O{P}_{Cu}}^o+E_{R{P}_{Ag}}^o)$
$E_{cell}=E_{cell}^o$
After the passage of 9.65 ampere for 1 hr, i.e., $9.65\times 60\times 60$ coulomb change, during which the cell reactions are reversed, thus, Ag metal passes in solution state and $C{u}^{2+}$ ions are discharged.
The reactions during this passage of current are:
$2Ag\to 2A{g}^{+}+2e$
$C{u}^{2+}+2e\to Cu$
Thus, $A{g}^{+}$ formed $=\frac{9.65\times 60\times 60}{96500}$
$=0.36eq.=0.36mole$
$C{u}^{2+}$ discharged $=\frac{9.65\times 60⋊60}{96500}$
$=0.36eq.=0.18mole$
Thus, $\left[A{g}^{+}\right]left=1+0.36=1.36mole$
$\left[C{u}^{2+}\right]left=1-0.18=0.82mole$
Now, e.m.f. can be given as
$E_{cell}=E_{cell}^o+\frac{0.059}{2}log\frac{(1.36{)}^2}{0.82}$
$=E_{cell}^o+0.010V$
Thus, $E_{cell}$ increases by $0.010V.$