Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the strengths of the oxidizing and reducing agents

$$\begin{gathered} {\mathbf{\text{(1)I}}}_{\mathbf{\text{2}}}{\mathbf{\text{(s)+ 2e}}}^{\mathbf{\text{-}}}{\mathbf{\text{n2I}}}^{\mathbf{\text{-}}}{\mathbf{\text{(aq)E}}}^{\mathbf{\text{0}}}\mathbf{\text{=0.53V}} \\ {\mathbf{\text{(2)S}}}_{\mathbf{\text{2}}}{{\mathbf{\text{O}}}_{\mathbf{\text{8}}}}^{\mathbf{\text{2 -}}}{\mathbf{\text{(aq) + 2e}}}^{\mathbf{\text{-}}}\mathbf{\text{n2S}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2 -}}}\mathbf{\text{(aq)}}{\mathit{E}}^{\mathbf{\text{0}}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{01}\mathit{V} \\ {\mathbf{\text{(3)Cr}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{7}}}^{\mathbf{\text{2 -}}}{\mathbf{\text{(aq) + 14H}}}^{\mathbf{\text{+}}}{\mathbf{\text{(aq) + 6e}}}^{\mathbf{\text{-}}}{\mathbf{\text{n2Cr}}}^{\mathbf{\text{3 +}}}{\mathbf{\text{(aq) + 7H}}}_{\mathbf{\text{2}}}{\mathbf{\text{O(l)E}}}^{\mathbf{\text{0}}}\mathbf{\text{=1.33V}} \end{gathered}$$

We have to write three spontaneous reactions from the given half-reactions:

$$\begin{gathered} {\text{(1)I}}_{\text{2}}{\text{(s)+ 2e}}^{\text{-}}{\text{n2I}}^{\text{-}}{\text{(aq)E}}^{\text{0}}\text{=0.53V} \\ {\text{(2)S}}_{\text{2}}{{\text{O}}_{\text{8}}}^{\text{2 -}}{\text{(aq) + 2e}}^{\text{-}}\text{n2S}{{\text{O}}_{\text{4}}}^{\text{2 -}}\text{(aq)}{E}^{\text{0}}=2.01V \\ {\text{(3)Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + 14H}}^{\text{+}}{\text{(aq) + 6e}}^{\text{-}}{\text{n2Cr}}^{\text{3 +}}{\text{(aq) + 7H}}_{\text{2}}{\text{O(l)E}}^{\text{0}}\text{=1.33V} \end{gathered}$$

For a reaction to be spontaneous,${\text{E}}_{\text{cel}}^{\text{c}}\text{> 0}$

(a) Combine (2) and reversed of (1):

${\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2 -}}{\text{(aq) + 2e}}^{\text{-}}{\text{+ 2I}}^{\text{-}}\text{(aq)}\to {\text{2SO}}_{\text{4}}^{\text{2 -}}{\text{(aq) +I}}_{\text{2}}\left(s\right){\text{+ 2e}}^{\text{-}}$

Cancel common terms to obtain the spontancous reaction:

$$\begin{gathered} {\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2}}{\text{(aq) + 2I}}^{\text{-}}{\text{2SO}}_{\text{4}}^{\text{2}}{\text{(aq) +I}}_{\text{2}}\left(s\right) \\ {E}_{cell}^0=E_{\text{red}}^C-E_{\text{cri}}^0\text{= 2.01 - 0.53 = 1.48V} \end{gathered}$$

(b) Combine $3\times \left(2\right)$and reversed of (3):

$3{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2}}{\text{(aq) + 6c + 2Cr}}^{\text{3}}{\text{+ (aq) + 7ZI}}_{\text{2}}\text{O(l)}\to {\text{6SO}}_{\text{4}}^{\text{2}}{\text{(aq) + Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2}}{\text{(aq) + 1III + (aq) + 6c}}^{\text{-}}$

Cancel common terms to obtain the spontaneons reaction:

$$\begin{gathered} 3{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2}}{\text{(aq) + 2Cr}}^{\text{3 +}}{\text{(aq) + 7H}}_{\text{2}}\text{O(I)}\to {\text{6SSO}}_{\text{4}}^{\text{2}}{\text{(aq) + Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2}}\text{(aq) + 14H + (aq)} \\ {E}_{cell}^0=E_{red}^0-E_{axi}^0=\text{2.01-1.33=0.68V} \end{gathered}$$

(c) Combine (3) and reversed of$3\times \left(1\right)$ :

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + 14H + (aq) + 6e}}^{\text{-}}{\text{+ 6I}}^{\text{-}}\left(\text{aq}\right)\to {\text{2Cr}}^{\text{3 +}}{\text{(aq)+7H}}_{\text{2}}{\text{O(l)+3I}}_{\text{2}}\left(s\right){\text{+6e}}^{\text{-}}$

Cancel common terms to obtain the spontaneous reaction:$$\begin{gathered} {\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}\left(\text{aq}\right){\text{+14H + (aq)+6I}}^{\text{-}}\text{(aq)}\to {\text{2Cr}}^{\text{3 +}}{\text{(aq)+7H}}_{\text{2}}{\text{O(l)+3I}}_{\text{2}}\left(S\right) \\ {E}_{CCl}^0=E_{red}^0-E_{aci}^0\text{= 1.33 - 0.53 = 0.8Vnendgathered} \end{gathered}$$

Oxidising agents :${\text{S}}_2{\text{O}}_8^{2-}\left(\text{aq}\right)>{\text{Cr}}_2{\text{O}}_7^{2-}\left(\text{aq}\right)>{\text{I}}_2(\text{s})$

Reducing agents:${\text{I}}^{-}\left(\text{aq}\right)>{\text{Cr}}^{3+}\left(\text{aq}\right)>{\text{SO}}_4^{2-}\left(\text{aq}\right)$

Hence (a)${\text{S}}_2{\text{O}}_8^{2-}\left(\text{aq}\right)+2{\text{I}}^{-}\left(\text{aq}\right)\to 2{\text{SO}}_4^{2-}\left(\text{aq}\right)+{\text{I}}_2(\text{s});E_{cell}^0=1.48\text{V}$

(b)$3{\text{S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2 -}}{\text{(aq) + 2Cr}}^{\text{3 +}}{\text{(aq) + 7H}}_{\text{2}}\text{O(l)}\to {\text{6SO}}_{\text{4}}^{\text{2 -}}{\text{(aq) + Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}\text{(aq) +}14\text{H}+\left(\text{aq}\right);E_{cell}^a=0.68\text{V}$

$$\begin{gathered} {\text{(c)Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + 14H + (aq) + 6I}}^{\text{-}}\text{(aq)}\to {\text{2Cr}}^{\text{3 +}}{\text{(aq)+7H}}_{\text{2}}{\text{O(l)+3I}}_{\text{2}}\left(S\right)\text{;} \\ {E}_{Cell}^0=0.8V \\ {\text{Oxidisingagents:S}}_{\text{2}}{\text{O}}_{\text{8}}^{\text{2 -}}{\text{(aq)>Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq)>I}}_{\text{2}}\left(s\right) \\ {\text{Reducingagents:I}}^{\text{-}}{\text{(aq) > Cr}}^{\text{3 +}}{\text{(aq) > SO}}_{\text{4}}^{\text{2 -}}\text{(aq)} \end{gathered}$$