Question

Explain the following with reasons:
(I)${\rm{(i)}}$Chlorine displaces iodine from ${\rm{KI}}$ solution but iodine does not displace bromine from ${\rm{KBr}}$ solution. Why?
(II)${\rm{(ii)}}$ ${\rm{Hg}} + {{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \to {\rm{HgS}}{{\rm{O}}_{\rm{4}}} + {{\rm{H}}_{\rm{2}}}$
Above reaction is not possible.

Answer 1

As we know that, the redox reaction is the reaction in which oxidation and reduction occurs simultaneously. The reaction occurs only when it is spontaneous. The reaction is spontaneous when reactant can oxidise or reduce to the other reactant.

Step by step answer: The spectrochemical series helps in finding out whether a given redox reaction is feasible or not from the ${{\rm{E}}^{\rm{0}}}$values of the two electrodes.
In general, a redox reaction is feasible only if the species which has higher reduction potential is reduced or accepts the electrons and the species which has lower reduction potential is oxidised or loses the electrons.
The oxidizing power of halogens decreases in the order as ${{\rm{F}}_{\rm{2}}}{\rm{ > C}}{{\rm{l}}_{\rm{2}}}{\rm{ > B}}{{\rm{r}}_{\rm{2}}}{\rm{ > }}{{\rm{I}}_{\rm{2}}}$. Therefore, ${{\rm{F}}_2}$can oxidize ${\rm{C}}{{\rm{l}}^{\rm{ - }}}$, ${\rm{B}}{{\rm{r}}^ - }$ and ${{\rm{I}}^ - }$ to respective ${\rm{C}}{{\rm{l}}_{\rm{2}}}$,${\rm{B}}{{\rm{r}}_{\rm{2}}}$and ${{\rm{I}}_{\rm{2}}}$,${\rm{C}}{{\rm{l}}_{\rm{2}}}$can oxidize ${\rm{B}}{{\rm{r}}^ - }$ and ${{\rm{I}}^ - }$ to respective${\rm{B}}{{\rm{r}}_{\rm{2}}}$and ${{\rm{I}}_{\rm{2}}}$ and ${\rm{B}}{{\rm{r}}_{\rm{2}}}$can oxidize ${{\rm{I}}^ - }$ to ${{\rm{I}}_{\rm{2}}}$.
In the given statement${\rm{(i)}}$ chlorine displaces iodine from ${\rm{KI}}$ solution because chlorine has more oxidizing power than iodine and oxidizes ${{\rm{I}}^ - }$ to ${{\rm{I}}_{\rm{2}}}$that’s why it displaces iodine from ${\rm{KI}}$. The reaction can be written as –
${\rm{C}}{{\rm{l}}_{\rm{2}}} + {\rm{KI}} \to {\rm{KCl}} + {{\rm{I}}_{\rm{2}}}$
But iodine is not an oxidizing agent so it cannot displace bromine from ${\rm{KBr}}$ solution.
${\rm{(ii)}}$ Mercury is the transition metal and it has more oxidizing power than hydrogen so it cannot reduce hydrogen. That’s why this reaction is not feasible.

Note: Some of the questions we get in which two reduction electrode potentials are given and we are unable to find out which is oxidation or which will occur on anode or cathode, in that case we should know that if the given reduction potential is positive or less negative then it will occur on anode and if the given reduction potential is more negative then it will occur on cathode.