1. ⇒ (MHT CET 2023 12th May Evening Shift )

Calculate the $E_{cell}$ for ${Zn}_{(s)} | {Zn}_{(0.1 M)}^{+ +} | | {Cr}_{(0.1 M)}^{+ + +} | {Cr}_{(s)}$ at $25^{\circ} C$ if $E_{cell}^{\circ}$ is $0.02 V$

A. $- 0.05 V$

B. $0.03 V$

C. $- 0.06 V$

D. $0.07 V$

Correct Option is (B)

The reaction is

$\begin{aligned} 3 {Zn}_{(3)} + 2 {Cr}_{(0.1 M)}^{3 +} ⟶ 3 {Zn}_{(0.1)}^{2 +} + 2 {Cr}_{(s)} \\ E_{cell} = E_{cell}^{o} - \frac{0.0592 V}{n} \log_{10} \frac{{[{Zn}^{2 +}]}^{3}}{{[{Cr}^{3 +}]}^{2}} \end{aligned}$

$\begin{aligned} ∴ E_{cell} & = 0.02 V - \frac{0.0592 V}{6} \log_{10} \frac{(0.1)^{3}}{(0.1)^{2}} \\ = 0.02 V - \frac{0.0592 V}{6} \log_{10} 0.1 \\ = 0.02 V + \frac{0.0592 V}{6} \times 1 \\ = 0.02 V + 0.0099 V \\ = 0.03 V \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Evening Shift )

Calculate the $E_{cell}^{\circ}$ for ${Zn}_{(s)} | {Zn}_{(IM)}^{+ +} | | {Cd}_{(IM)}^{+ +} | {Cd}_{(s)}$ at $25^{\circ} C [E_{Zn}^{0} = - 0.763 V; E_{Cd}^{\circ} = - 0.403 V]$

A. $0.36 V$

B. $1.17 V$

C. $- 0.36 V$

D. $- 1.17 V$

Correct Option is (A)

For the given cell reaction, anode is $Zn$ and cathode is $Cd$ .

$\begin{aligned} E_{cell}^{0} & = E_{cathode}^{0} - E_{anode}^{0} \\ = - 0.403 - (- 0.763) \\ = 0.36 V \end{aligned}$

3. ⇒ (MHT CET 2023 10th May Evening Shift )

Calculate $E_{cell}^{\circ}$ in which following reaction occurs. ${Mg}_{(s)} + 2 {Ag}_{(IM)}^{+} \to {Mg}_{(1 M)}^{+} + 2 {Ag}_{(s)}$ if $E_{Ag}^{\circ} = 0.8 V$ and $E_{Mg}^{\circ} = - 2.37 V$

A. $- 3.17 V$

B. $3.17 V$

C. $- 1.57 V$

D. $1.57 V$

Correct Option is (B)

For the given cell reaction, anode is $Mg$ and cathode is Ag.

$\begin{aligned} E_{edl}^{o} & = E_{cuthode}^{\circ} - E_{unode}^{\circ} \\ = 0.8 - (- 2.37) \\ = 3.17 V \end{aligned}$

4. ⇒ (MHT CET 2023 9th May Evening Shift )

A reaction, ${Ni}_{(s)} + {Cu}_{((M))}^{+} \to {Ni}_{(IM)}^{+} + {Cu}_{(s)}$ occurs in a cell. Calculate $E_{cell}^{\circ}$ if $E_{Cu}^{\circ} = 0.337 V$ and $E_{Ni}^{\circ} = - 0.257 V$

A. $0.594 V$

B. $- 0.594 V$

C. $- 0.08 V$

D. $0.08 V$

Correct Option is (A)

The standard cell potential is given by

$\begin{aligned} E_{cell}^{\circ} & = E_{cathode}^{\circ} - E_{anode}^{\circ} \\ E_{cell}^{\circ} & = E_{Cu}^{\circ} - E_{Ni}^{\circ} \\ = (0.337 V) - (- 0.257 V) \\ = 0.337 V + 0.257 V \\ = 0.594 V \end{aligned}$

⇒Electrochemistry