${{\mathrm{E}}^{⊝}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}=0.74 \mathrm{V}$

${{\mathrm{E}}^{⊝}}_{{\mathrm{Cd}}^{2+}/\mathrm{Cd}}=-0.40 \mathrm{V}$

The galvanic cell of the given reaction is depicted as:

${\mathrm{Cr}}_{\left(\mathrm{s}\right)}\left|{{\mathrm{Cr}}^{3+}}_{\left(\mathrm{aq}\right)}\right|\left|{{\mathrm{Cd}}^{2+}}_{\left(\mathrm{aq}\right)}\right|{\mathrm{Cd}}_{\left(\mathrm{s}\right)}$

Now, the standard cell potential is

${\mathrm{E}}_{\mathrm{cell}}^{⊝}={\mathrm{E}}_{\mathrm{R}}^{⊝}-{\mathrm{E}}_{\mathrm{L}}^{⊝}$

       = -0.40-(-0.74)

       =+0.34 V

Step 2:

Calculate the value of \(\Delta_{r} G^{o} \) is as follows:

${∆}_{\mathrm{r}}{\mathrm{G}}^{⊝}=-{\mathrm{nFE}}_{\mathrm{cell}}^{⊝}$

In the given equation, n=6

F=96487 C ${\mathrm{mol}}^{-1}$  

${\mathrm{E}}_{\mathrm{cell}}^{⊝}$=+0.34 V

Then, ${∆}_{\mathrm{r}}{\mathrm{G}}^{⊝}$= -6 x 96487 C ${\mathrm{mol}}^{-1}$ x 0.34 V

=-196833.48 CV mol^-1

=-196833.48 J mol^-1

=-196.83 kJ mol^-1

Step 3:

Again,

${∆}_{\mathrm{t}}{\mathrm{G}}^{⊝}=-\mathrm{RT} \mathrm{In} \mathrm{K}$

$\Rightarrow {∆}_{\mathrm{t}}{\mathrm{G}}^{⊝}=-2.303 \mathrm{RT} \mathrm{In} \mathrm{K}$

$\Rightarrow \log \mathrm{K}=-\frac{{∆}_{\mathrm{t}}\mathrm{G}}{2.303 \mathrm{RT}}$

$=\frac{-196.83\times {10}^3}{2.303\times 8.314\times 298}$

$=34.496$

$\therefore \mathrm{K}=\mathrm{antilg} (34.496)=$$3.13\times {10}^{34}$
 

(ii) 

Step 2:

${\mathrm{First\; calculate\; the\; value\; of }}^{}$\(E_{cell}^{o}\)${\mathrm{}}^{}$ is as follows:

${{\mathrm{E}}^{⊝}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.77 \mathrm{V}$

${{\mathrm{E}}^{⊝}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=0.80 \mathrm{V}$

The galvanic cell of the given reaction is depicted as:

${{\mathrm{Fe}}^{2+}}_{\left(\mathrm{aq}\right)}\left|{{\mathrm{Fe}}^{3+}}_{\left(\mathrm{aq}\right)}\right|\left|{{\mathrm{Ag}}^{+}}_{\left(\mathrm{aq}\right)}\right|{\mathrm{Ag}}_{\left(\mathrm{s}\right)}$

Now, the standard cell potential is

${\mathrm{E}}_{\mathrm{cell}}^{⊝}={\mathrm{E}}_{\mathrm{R}}^{⊝}-{\mathrm{E}}_{\mathrm{L}}^{⊝}$

$=0.80-0.77$

$=0.03 \mathrm{V}$

Here, n=1

Step 2:

Then, ${∆}_{\mathrm{t}}{\mathrm{G}}^{⊝}=-{\mathrm{nFE}}_{\mathrm{cell}}^{⊝}$

$=-1\times 96487 \mathrm{C} {\mathrm{mol}}^{-1} \times 0.03 \mathrm{V}$

$=-2894.61 \mathrm{J} {\mathrm{mol}}^{-1}$

$=\; -2.89\; kJ\; /mol\mathrm{ }$

Step 3:

$\mathrm{Again}, {∆}_{\mathrm{t}}{\mathrm{G}}^{⊝}=-2.303 \mathrm{RT} \mathrm{In} \mathrm{K}$

$\Rightarrow \log \mathrm{K}=-\frac{{∆}_{\mathrm{t}}\mathrm{G}}{2.303 \mathrm{RT}}$

$=\frac{-2894.61}{2.303\times 8.314\times 298}$

$=0.5073$

$\mathrm{K}=\mathrm{antilog} (0.5073)$

=3.2 (approximately)