Question

For the given cell at $298K$, calculate the value of cell potential:

$Ag\left(s\right)|A{g}^{+}(aq,0.02M)||A{g}^{+}(aq,2.0M)|Ag\left(s\right)$

• A. $-0.059V$
• B. $+0.059V$
• C. $+0.118V$
• D. $-0.118V$

Answer 1

The correct option is C $+0.118V$

A galvanic cell which ​consists of both electrodes ​of the same material​ dipped into the same electrolytic solution.​ The electrolytic solution will be at different concentrations. These two solutions are separated by a salt bridge. ​

Here, the silver electrode is dipped in an aqueous solutions having different concentration of $0.02M$ and $2M$

In concentration cell, for a spontaneous process, the electrode with higher concentration of electrolyte will act as cathode and lower concentration half cell will act as anode.

For the following cell :
$Ag\left(s\right)|A{g}^{+}(aq,0.02M)||A{g}^{+}(aq,2.0M)|Ag\left(s\right)$

The electrode half cell reaction will be
$A{g}^{+}(aq,2M)+e^{-}\to Ag\left(s\right)$
$Ag\left(s\right)\to A{g}^{+}(aq,0.02M)+e^{-}$

The cell reaction will be
$A{g}^{+}(aq,2M)\to A{g}^{+}(aq,0.02M)$

By Nernst equation,

$E_{cell}=E^0-\frac{0.059}{1}log\frac{0.02}{2.0}$

For concentration cells,
$E^0=0$

$E_{cell}=-\frac{0.059}{1}log\frac{0.02}{2.0}$

$E_{cell}=-\frac{0.059}{1}log0.01$

$E_{cell}=\frac{0.059}{1}\times 2$

$E_{cell}=0.118V$