Q21.49 P - Chemistry: Molecular Nature Of Matter And Change

Question

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 

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

Step-by-Step Solution

Answer

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

 

$14\text{H}+\left(\text{aq}\right);E_{cell}^a=0.68 \text{V}$

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

Step 1: To write three spontaneous reactions

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 $3\times \left(1\right)$ :

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

Step 2: To Cancel common terms

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

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

 

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