Starting with correctly balanced half reactions, write the overall net ionic reaction in the following changes:
a. Chloride ion is oxidised to `Cl_(2)` by `underline(Mn)O_(4)^(Ө)` in acid solution.
b. Nitrous acid `(Hunderline(N)O_(2))` reduce `MnO_(4)^(Ө)` in acidsolution.
c. Nitrous acid `(Hunderline(N)O_(2))` oxidises `I^(Ө)` to `I_(2)` in acid solution.
d. Chlorate ion `(underline(Cl)O_(3)^(Ө))` oxidises `Mn^(2+)` to `MnO_(2)(s)` in acid solution.
e. Chromine ion `(underline(C )rO_(3)^(Ө))` is oxidation numbers of the basic solution.
Also find out the change in oxidation numbers of the underline atoms.

Answer

A

Explanation

i. `2MnO_(4)^(ө)+16H^(o+)+10CI^(ө)rarr5Cl_(2)+2Mn^(2+)+8H_(2)O`
ii. `2MnO_(4)^(ө)+6H^(+)+5NO_(2)^(ө)rarr5NO_(3)^(ө)+3H_(2)O+2Mn^(2+)`
iii. `2HNO_(2)+2H^(o+)+2I^(ө)rarr2NO+2H_(2)O+I_(2)`
IV. `ClO_(3)^(ө)+3Mn^(2+)+3H_(2)Orarr3MnO_(2)+Cl^(ө)+6H^(o+)`
v. `2CrO_(3)^(ө)+2overset(ө)OH+H_(2)O_(2)rarr2CrO_(4)^(2-)+2H_(2)O`
Oxidation numbers:
a. Oxidation number of `Mn` changes from `+7` to `+2`
b. Oxidation number of `N` changes from `+3` to `+5`.
c. Oxidation number of `N` changes from `+3` to `+2`.
d. Oxidation number of `N` changes from `+5` to `-1`
e. Oxidation number of `N` changes from `+5` to` +6`.