Question

Given:
Pt (s) | $C{l}^{-}$($C_1$) | $C{l}_2$(g)($P_1$ atm) || $C{l}_2$(g)($P_1$ atm)| $C{l}^{-}$($C_2$)| Pt (s)
Identify in which of the following condition working of cell takes place:

• A. $C_1$ > $C_2$ and $P_1$ = $P_2$
• B. $P_2$ > $P_1$ and $C_1$ = $C_2$
• C. $C_1$ < $C_2$ and $P_1$ = $P_2$
• D. $P_2$ < $P_1$ and $C_1$ = $C_2$

Answer 1

Answer: (A), (B)

$P_2$ > $P_1$ and $C_1$ = $C_2$

$2\underset{\left(C_{1}M\right)}{C{l}^{-}}\to \underset{P_{1}atm}{C{l}_2}+2e^{-}$$\underset{P_2\mathrm{atm}}{{\mathrm{Cl}}_2}+2e^{-}\to 2\underset{\left(C_{2}M\right)}{{\mathrm{Cl}}^{-}}$
On adding above 2 equations we get,
$2\underset{\left(C_{1}M\right)}{{\mathrm{Cl}}^{-}}+\underset{P_2\mathrm{atm}}{{\mathrm{Cl}}_2}\to \underset{P_1\mathrm{atm}}{{\mathrm{Cl}}_2}+2\underset{\left(C_{2}M\right)}{{\mathrm{Cl}}^{-}}$
$E_{\mathrm{cell}}=E{°}_{\mathrm{cell}}-\frac{0.0591}{2}\log \frac{P_1\times C_2^2}{P_2\times C_1^2}$
As this is the concentration cell so $E{°}_{\mathrm{Cell}}=0$
$E_{\mathrm{Cell}}=-\frac{0.0591}{2}\log \frac{P_1\times {C_2}^2}{P_2\times {C_1}^2}$
To make the reaction spontaneous $E_{\mathrm{Cell}}>0$
So, option (a) and (b) are correct