91Ó°ÊÓ

For a redox reaction taking place, the standard reduction potential is the difference between the respective cell potential.

${\text{E}}_{\text{cell}}°={\text{E}}_{\text{cathode}}°{\text{- E}}_{\text{anode}}°$

The relation between and the standard electrode potential is given below.

$△G^{∘}=-nF{E^{∘}}_{cell}$

Where,

n = number of electrons involved in the redox reaction

$\text{F} \text{=} \text{96500} \text{C/mol}$.

The redox reaction taking place between $\text{Ag}\left(\text{s}\right)  \text{and}  {\text{Mn}}^{\text{+ 2}}\left(\text{aq}\right)$ is given below.

$2\text{Ag}\left(\text{s}\right)+{\text{Mn}}^{2+}→2{\text{Ag}}^{+}\left(\text{aq}\right)+\text{Mn}\left(\text{s}\right)$

Now, we have to write down the cathode and anode half-cell reaction along with their respective electrode potential.

$$\begin{gathered} \text{2Ag}\left(\text{s}\right)→{\text{2Ag}}^{\text{+}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}                      {{\text{E}}^{\text{0}}}_{\text{anode}}=0.80 \text{V} \\ {\text{Mn}}^{\text{+ 2}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}→\text{Mn}\left(\text{s}\right)                         {{\text{E}}^{\text{0}}}_{\text{cathode}}=-1.18 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{∘}}_{cell}={E^{∘}}_{Cathode}-{E^{∘}}_{anode} \\ {E^{∘}}_{cell}=-1.18V-0.80V \\ {E^{∘}}_{cell}=-1.98V \end{gathered}$$

We know that,

$$\begin{gathered} △G^{∘}=-nF{E^{∘}}_{cell} \\ △G^{∘}=-2×96500×-1.98 \\ △G^{∘}=382.14KJ/mol \end{gathered}$$

Hence$△G^{∘}=382.14KJ/mol$.

The redox reaction taking place between ${\text{l}}_2\left(\text{s}\right)  \text{and}  {\text{Br}}^{\text{-}}\left(\text{aq}\right)$is given below.

${\text{Cl}}_2(\text{s})+2{\text{Br}}^{-}\left(\text{aq}\right)→2{\text{Cl}}^{-}\left(\text{aq}\right)+{\text{Br}}_2\left(\text{l}\right)$

Now, we have to write down the cathode and anode half-cell reaction along with their respective electrode potential.

$$\begin{gathered} {\text{Cl}}_{\text{2}}\left(\text{s}\right){\text{+ 2e}}^{\text{-}}→{\text{2Cl}}^{\text{-}}\left(\text{aq}\right)                        {{\text{E}}^{\text{0}}}_{\text{cathode}}=1.36\text{V} \\ {\text{2Br}}^{\text{-}}\left(\text{aq}\right)    →{\text{Br}}_{\text{2}}\left(\text{s}\right){\text{+ 2e}}^{\text{-}}                     {{\text{E}}^{\text{0}}}_{\text{anode}}=1.07 \text{V} \end{gathered}$$

$$\begin{gathered} {E^{∘}}_{cell}={E^{∘}}_{reduction}-{E^{∘}}_{oxidation} \\ {E^{∘}}_{cell}=\left(1.36-1.07\right)V \\ {E^{∘}}_{cell}=-0.29V \end{gathered}$$

We know that.

$$\begin{gathered} △G^{∘}=-nF{E^{∘}}_{cell} \\ △G^{∘}=-2.96500C/mol×0.29V \\ △G^{∘}=-55.97KJ/mol \end{gathered}$$

hence $△G^{∘}=-55.97KJ/mol$