Question

For the given cell ;
$Cu(s)\left|C{u}^{2+}\left(C_{1}M\right)\Vert C{u}^{2+}\left(C_{2}M\right)\right|Cu(s)$ change in Gibbs energy $(\Delta G)$ is negative, if

• A. $C_1=2C_2$
• B. $C_2=C_1/\sqrt{2}$
• C. $C_1=C_2$
• D. $C_2=\sqrt{2}{C}_1$

Answer 1

Answer: (D)

$\Delta G=-nF{E}_{\text{cell}}$

$\Delta G$ is negative, if $E_{\text{cell}}$ is positive

Anode: $Cu(s)\to C{u}^{+2}(C_1)+2e^{-}:E^{\circ }$

$\frac{\text{ Cathode }:C{u}^{+2}\left(C_2\right)+2e^{-}\to Cu(S):-E^{\circ }}{\text{ Cell reaction }:C{u}^{+2}\left(C_2\right)⟶C{u}^{+2}\left(C_1\right)E_{cen}^{\circ }=0}$

$E_{\text{cell }}=E_{cell}^{\circ }-\frac{2.303RT}{nF}\log Q$

$E_{\text{cell }}=0-\frac{2.303RT}{nF}\log \left(\frac{C_1}{C_2}\right)$

$E_{\text{cell }}>0$:

if $\frac{C_1}{C_2}<1$

$\Rightarrow C_1<C_2$