Question

Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3no2− + 8h+ + cr2o72− → 3no3− +2cr3+ + 4h2o classify each reactant as the reducing agent, oxidizing agent, or neither.

Answer 1

Answer:-
`NO_2^-` is a reducing agent,
`Cr_2O_7^2^-` is an oxidizing agent and
`H^+` is neither.

Solution:- We write the oxidation numbers for all the elements and see for which of them it's changing. An increase in oxidation number is oxidation and a decrease in oxidation number is reduction. What is being oxidized acts as a reducing agent and what is being reduced acts as an oxidizing agent.

Oxidation number of N in
`NO_2^-` is +3 and O is -2 . Oxidation number of H in
`H^+` is +1 . Oxidation number of Cr in
`Cr_2O_7^2^-` is +6 and O is -2 .

Oxidation number of N in
`NO_3^-` is +5 and O is -2. Oxidation number of Cr in
`Cr^3^+` is +3. oxidaton numbers of H and O in water are +1 and -2 respectively.

Looking at all of these only the oxidation number of N is increasing from +3 to +5 which indicates oxidation and so
`NO_2^-` is a reducing agent.

Oxidation number of Cr is decreasing from +6 to +3 which indicates reduction and so
`Cr_2O_7^2^-` is an oxidizing agent.

Oxidation numbers of H and O are not changing and hence
`H^+` is neither oxidizing nor reducing agent.