Solution:
(i) Given: 
![\[E_{C d^{2+} / C d}^{\Theta}=-0.40 V\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-19ccf6b70e64b62e9add3a2525c118c7_l3.png "Rendered by QuickLaTeX.com")
The galvanic cell of the given reaction is represented as :
![\[C r_{(s)}\left|C r_{(a q)}^{3+} | C d_{a q}^{2+}\right| C d_{(s)}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-085a569c5cb6bfce9f37a55bde5ebc72_l3.png "Rendered by QuickLaTeX.com")
Now, the standard cell potential is
![\[\begin{aligned}&E_{c e l l}^{\Theta}=E_{g}^{\Theta}-E_{L}^{\Theta} \&=-0.40-(-0.74) \&=+0.34 V\end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ff37daad0eeda561ed278c371396eb43_l3.png "Rendered by QuickLaTeX.com")
![\[E_{c e l l}^{\Theta}=+0.34 V\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b00b4ec4dacb1ed9253011db58db1156_l3.png "Rendered by QuickLaTeX.com")
In the given equation,
![\[F =96487 C mol ^{-1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-26848cc034255104334afb91bc32038f_l3.png "Rendered by QuickLaTeX.com")
Then substuiting the value of parameters in the formula we get,
![\[\begin{aligned}&=-196833.48 CV mol -1 \&=-196833.48 J mol -1 \&=-196.83 kJ mol -1\end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-70bc97c5f1b63dc00582b6d7fcf8e7f3_l3.png "Rendered by QuickLaTeX.com")
![\[\begin{aligned}&\Delta_{r} G^{\Theta}=-R T l n K \quad \Delta_{r} G^{\Theta}=-2.303 R T l n K \log k=\frac{\Delta_{r} G}{2.303 R T}=\frac{-196.83 \times 10^{3}}{2.303 \times 8.314 \times 298} \&=34.496 \& K =\operatorname{antilog}(34.496)=3.13 \times 10^{34}\end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-353f366ffbd44dd6d9a501fa889362f0_l3.png "Rendered by QuickLaTeX.com")
ii) The galvanic cell of the given reaction is represented as follows:
![\[F e_{(a q)}^{2+}\left|F e_{(a q)}^{3+} | A g_{(a q)}^{+}\right| A g_{(s)}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-054e928b1d647e64f33e49fcf342f76d_l3.png "Rendered by QuickLaTeX.com")
Now, the standard cell potential is
![\[E_{c e l l}^{\Theta}=E_{g}^{\Theta}-E_{L}^{\Theta}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0687082800df1976e08283899ebe4113_l3.png "Rendered by QuickLaTeX.com")
Here,
Then substuiting we have,
![\[\begin{aligned}&=-1 \times 96487 C mol ^{-1} \times 0.03 V \&=-2894.61 J mol ^{-1} \&=-2.89 kJ mol ^{-1}\end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cfde1759eb756fcd4094d21eabad5c0e_l3.png "Rendered by QuickLaTeX.com")
Again,
