Question
At $$298\,K,$$ the standard reduction potentials are $$1.51\,V$$ for $$MnO_4^ - \left| {M{n^{2 + }},1.36\,V\,{\text{for}}\,C{l_2}} \right|$$ $$C{l^ - },1.07V\,{\text{for}}\,B{r_2}$$ $$\left| {B{r^ - },{\text{and}}\,0.54V\,} \right.{\text{for}}\,{I_2}\left| {{I^ - }.} \right.$$ $$At\,pH = 3,$$ permanganate is expected to oxidize : $$\left( {\frac{{RT}}{F} = 0.059V} \right)$$
A.
$$C{l^ - },B{r^ - }\,{\text{and}}\,{I^ - }$$
B.
$$B{r^ - }\,{\text{and}}\,{I^ - }$$
C.
$$C{l^ - }\,{\text{and}}\,B{r^ - }$$
D.
$${I^ - }\,{\text{only}}$$
Answer :
$$B{r^ - }\,{\text{and}}\,{I^ - }$$
Solution :
$$\eqalign{
& MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O \cr
& E = 1.51 - \frac{{0.059}}{5}\log \frac{{\left[ {M{n^{2 + }}} \right]}}{{\left[ {MnO_4^ - } \right]{{\left[ {{H^ + }} \right]}^8}}} \cr} $$
Taking $$M{n^{2 + }}$$ and $$MnO_4^ - $$ in standard state i.e. $$1\,M,$$
$$\eqalign{
& E = 1.51 - \frac{{0.059}}{5} \times 8\log \frac{1}{{\left[ {{H^ + }} \right]}} \cr
& = 1.51 - \frac{{0.059}}{5} \times 8 \times 3 \cr
& = 1.2268\,V \cr} $$
Hence at this $$pH,MnO_4^ - $$ will oxidise only $$B{r^ - }$$ and $${I^ - }$$ as $$SRP$$ of $$\frac{{C{l_2}}}{{C{l^ - }}}$$ is $$1.36\,V$$ which is greater than that for $$\frac{{MnO_4^ - }}{{M{n^{2 + }}}}.$$