Question

For the given cell:

$$\begin{gathered} \mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\right({\mathrm{C}}_1\mathrm{M}\left)\right|\left|{\mathrm{Cu}}^{2+}\right({\mathrm{C}}_2\mathrm{M}\left)\right|\mathrm{Cu}\left(\mathrm{s}\right) \\ \mathrm{Change}\mathrm{in}\mathrm{Gibbs}\mathrm{energy}\left(\mathrm{\Delta G}\right)\mathrm{is}\mathrm{negative},\mathrm{if}: \end{gathered}$$

• A. ${\mathrm{C}}_2=\sqrt{2}{\mathrm{C}}_1$
• B. ${\mathrm{C}}_2={\mathrm{C}}_1\sqrt{2}$
• C. ${\mathrm{C}}_1=2{\mathrm{C}}_2$
• D. ${\mathrm{C}}_1={\mathrm{C}}_2$

Answer 1

The correct option is A

${\mathrm{C}}_2=\sqrt{2}{\mathrm{C}}_1$

The explanation for the correct options:

A. The Gibbs free energy will be negative when${\mathrm{C}}_2=\sqrt{2}{\mathrm{C}}_1$

Step1: Given:$\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\right({\mathrm{C}}_1\mathrm{M}\left)\right|\left|{\mathrm{Cu}}^{2+}\right({\mathrm{C}}_2\mathrm{M}\left)\right|\mathrm{Cu}\left(\mathrm{s}\right)$ with$∆\mathrm{G}=-\mathrm{ve}$,$\mathrm{R}=8.303{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}$, $\mathrm{T}=248\mathrm{K}$,$\mathrm{f}=96500$

Step2: Formula: ${\mathrm{E}}_{\mathrm{cell}}=\mathrm{E}{°}_{\mathrm{cell}}-2.303\mathrm{RT}/\mathrm{nf}\log [{\mathrm{C}}_1/{\mathrm{C}}_2]$

Step 3:

$$\begin{gathered} \mathrm{Anode}:\mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{2+}\left({\mathrm{C}}_2\mathrm{M}\right) \\ \mathrm{Cathode}:{\mathrm{Cu}}^{2+}\left({\mathrm{C}}_1\mathrm{M}\right)\to \mathrm{Cu}\left(\mathrm{s}\right)+2{\mathrm{e}}^{-} \\ \mathrm{Overall}:{\mathrm{Cu}}^{2+}\left({\mathrm{C}}_1\mathrm{M}\right)\to {\mathrm{Cu}}^{2+}\left({\mathrm{C}}_2\mathrm{M}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=\mathrm{E}{°}_{\mathrm{cell}}-2.303\mathrm{RT}/\mathrm{nf}\log [{\mathrm{C}}_2/{\mathrm{C}}_1] \\ {\mathrm{E}}_{\mathrm{cell}}=0-0.0591/2\log [{\mathrm{C}}_2/{\mathrm{C}}_1] \\ {\mathrm{E}}_{\mathrm{cell}}=-0.0591/2\log [{\mathrm{C}}_2/{\mathrm{C}}_1] \\ \log [\mathrm{a}/\mathrm{b}]=\mathrm{loga}-\mathrm{logb} \\ {\mathrm{E}}_{\mathrm{cell}}=-0.0591/2{\mathrm{logC}}_2-{\mathrm{logC}}_1 \\ \mathrm{For}∆\mathrm{G}\mathrm{negative},\mathrm{the}{\mathrm{E}}_{\mathrm{cell}}\mathrm{should}\mathrm{be}\mathrm{positive}.\mathrm{So},{\mathrm{logC}}_2-{\mathrm{logC}}_1\mathrm{will}\mathrm{be}\mathrm{negative}. \end{gathered}$$

The explanation for the incorrect options:

B.${\mathrm{C}}_2={\mathrm{C}}_1\sqrt{2}$

Here ${\mathrm{C}}_2/{\mathrm{C}}_1=\sqrt{2}$, The concentration of ${\mathrm{C}}_1$is smaller than ${\mathrm{C}}_2$. Due to this, the value of $∆\mathrm{G}$ will become positive.

C.${\mathrm{C}}_1=2{\mathrm{C}}_2$

The concentration of ${\mathrm{C}}_2$is greater than ${\mathrm{C}}_1$. Due to this ${\mathrm{E}}_{\mathrm{cell}}$will become negative and the value of $∆\mathrm{G}$ will be positive.

D.${\mathrm{C}}_1={\mathrm{C}}_2$

If both the concentrations will be equal ${\mathrm{E}}_{\mathrm{cell}}=-0.0591/2\log [{\mathrm{C}}_1/{\mathrm{C}}_1]$, then ${\mathrm{E}}_{\mathrm{cell}}$will become negative and the value of $∆\mathrm{G}$ will be positive.

Thus, the correct option is A ${\mathrm{C}}_1=\sqrt{2}{\mathrm{C}}_2$.