Question

Two concentration cells of $Ag$ with $Ag$ electrode in $AgN{O}_3$ . In first cell concentration of one electrode is $1M$ and other electrode is $M$ and emf is $0.06V$. In second cell concentration of one electrode is $1M$ and other electrode is $0.01M$, calculate emf of second cell.

• A. 0.12V
• B. 0.06V
• C. 0.09V
• D. 0.16V

Answer 1

Answer: (A)

$E_{\text{cell }}=E^o-\frac{0.059}{n}\log \frac{[\text{ Product conc. }]}{[\text{ Reactant cone. }]}$
${\left(E_{\text{cell }}\right)}_1=0-\frac{0.059}{1}\log \frac{0.1}{1}\dots$ (i)
${\left(E_{\text{cell }}\right)}_2=0-\frac{0.059}{1}\log \frac{0.01}{1}\dots$ (ii)
$0.06=-\frac{0.059}{1}\log \frac{1}{10}\dots$ (i)
${\left(E_{\text{cell }}\right)}_2=-\frac{0.059}{1}\log \frac{1}{10}\dots$ (ii)
Dividing Eq. (i) by (ii),
we have $\frac{0.06}{(E_{\text{cell }{)}_2}}=\frac{\log 1-\log 10}{\log 1-\log 100}\frac{0.06}{(E_{\text{cell }{)}_2}}=\frac{0-1}{0-2\log 10}$
$\frac{0.06}{(E_{\text{cell }{)}_2}}=\frac{-1}{-2}=\frac{1}{2}{\left(E_{\text{cell }}\right)}_2=2\times 0.06=0.12V$