Write balanced redox reactions for each of the following reactions:
(a) Potassium dichromate `(K_(2)Cr_(2)O_(7))` reacts with hydroiodic acid `(HI)` to produce potassium iodide, chromium (III) iodide, and solid iodine, `I_(2)(s)`.
(b) A purple solution of aqueous potassium permanganate `(KMnO_(4))` reacts with aqueous sodium sulphite `(Na_(2)SO_(3))` in basic solution to yield the green magnanate ion `(MnO_(4)^(2-))` and sulphate ion `(SO_(4)^(2-))`.
(c ) `Sn^(2+) (aq)` reduce `I_(4)^(Ө)(aq)` to `I^(Ө)(aq)` and is oxidised to `Sn^(4+)`.
(d) `H_(2)O_(2)(aq)` oxidises `Mn^(2+)(aq)` to `MnO_(2)` in basic medium.
(e) `H_(2)O_(2)(aq)` reduces `Cr_(2)O_(7)^(2-) (aq)` to green coloured `Cr^(3+)(aq)` in acidic medium.

Answer

A

Explanation

(a) `{:(Cr_(2)O_(7)^(2-)+cancel(6e^(-))+14H^(o+)rarr2Cr^(3+)+7H_(2)O),(2I^(ө)rarrI_(2)+cancel(2e^(-)]xx3),(ulbar(Cr_(2)O_(7)^(2-)+6I^(ө)+14H^(o+)rarr2Cr^(3+)+3I_(2)+7H_(2)O)):}`
Add other ions to both sides, i.e.,
`LHS[[2K^(o+),],[81^(ө),]] RHS[[2K^(o+),],[81^(ө),]]`
The next redox equation is:
`K_(2)Cr_(2)O_(7)+14HIrarr2CrI_(3)+3KI+3I_(2)+7H_(2)O`
b. `cancel(e^(-))+MnO_(4)^(ө)rarrMnO_(4)^(2-)]xx2`
`x-8 = -1, x-8 = -2`
`x=7, x=6`
`2overset(ө)OH+cancel(H_(2)O)+SO_(3)^(2-)rarrSO_(4)^(2-)+cancel(2e^(-))+cancel(2)H_(2)O`
`x-6=-2`, `x-8 = -2`
`x=4, x=6`
`ulbar(2MnO_(4)^(ө)+SO_(3)^(2-)+2overset(ө)OHrarr2MnO_(4)^(2-)+SO_(4)^(2-)+H_(2)O)`
Adding other ions to both sides, i.e., `4K^(o+), 2Na^(o+)`, the net redox equation is
`ul(2KMnO_(4)+Na_(2)SO_(3)+2KOHrarr2K_(2)MnO_(4)+Na_(2)SO_(4)+H_(2)O)`
c.