Question

Two concentration cells of $Ag$ with $Ag$ electrode in $ AgNO_{3} $ . In first cell concentration of one electrode is $1 \,M$ and other electrode is $M$ and emf is $0.06\, V$. In second cell concentration of one electrode is $1 \,M$ and other electrode is $0.01\, M$, 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. }]} $
$\left(E_{\text {cell }}\right)_{1}=0-\frac{0.059}{1} \log \frac{0.1}{1} \ldots$ (i)
$\left(E_{\text {cell }}\right)_{2}=0-\frac{0.059}{1} \log \frac{0.01}{1} \ldots $ (ii)
$0.06=-\frac{0.059}{1} \log \frac{1}{10} \ldots $ (i)
$\left(E_{\text {cell }}\right)_{2}=-\frac{0.059}{1} \log \frac{1}{10} \ldots $ (ii)
Dividing Eq. (i) by (ii),
we have $ \frac{0.06}{\left(E_{\text {cell })_{2}}\right.}=\frac{\log 1-\log 10}{\log 1-\log 100} \frac{0.06}{\left(E_{\text {cell })_{2}}\right.}=\frac{0-1}{0-2 \log 10}$
$\frac{0.06}{\left(E_{\text {cell })_{2}}\right.}=\frac{-1}{-2}=\frac{1}{2}\left(E_{\text {cell }}\right)_{2}=2 \times 0.06=0.12 \,V$