Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron (II) solution and (iii) `H_(2)S`

Explanation

In acidic medium,`K_(2)Cr_(2)O_(7)` is a powerful oxidizing agent. The oxidation state of chromium changes from (+6) to (+3)
`Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) rarr 2Cr^(3+) + 7H_(2)O`
(i) Iodide: Iodide `(I^(-))` is oxidized to `I_(2)`. Reduction half reaction : `6I^(-) rarr I_(2) + 6e^(-)`
Reduction half reaction: `Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) rarr 2Cr^(3+) + 7H_(2)O`
Net reaction: `Cr_(2)O_(7)^(2-) + 6I_(-) + 14H^(+) rarr I_(2) + 2Cr^(3+) + 7H_(2)O`
(ii) Iron (II) solution: `Fe^(2+)` salts get oxidized to `Fe^(3+)` salts.
Oxidation half reaction: `6I^(-) rarr I_(2) + 6e^(-)`
Reduction half reaction: `Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) rarr 2Cr^(3+) + 7H_(2)O`
Oxidation half reaction : `6Fe^(2+) rarr 6Fe^(3+) + 6e^(-)` Net reaction:
`Cr_(2)O_(7)^(2-) + 6Fe^(2+) + 14H^(+) rarr 2Cr^(3+) + 6Fe^(3+) + 7H_(2)O`
(iii) `H_(2)S`: `H_(2)S` is oxidizd to sulphur Reduction half reaction: `Cr_(2)O_(7)^(2-) + 14H^(+) + 6e^(-) rarr 2Cr^(3+) + 7H_(2)O`
Oxidation half reactin : `3H_(2)S rarr 6H^(+) + 3S + 6e^(-)` Not reaction: `3H_(2)S + 8H^(+) + Cr_(2)O_(7)^(2-) rarr 2Cr^(3+) + 3S + 7H_(2)O`