Question

Identify the oxidizing agent and the reducing agent in the following: (i)8H+(aq) + [Cr2O7]2-(aq) + 3SO32-(aq) => 2Cr3+(aq) + 3SO42-(aq) + 4H2O(l) (ii) NO3-(aq) + 4Zn(s) + 7OH-(aq) + 6H2O(l) => 4Zn(OH)42-(aq) + NH3(aq) answer is (i) ox. agent = [Cr2O7]2-, red. agent = SO32- (ii) ox. agent = NO3-, red agent = Zn

Answer 1

Answer – in this we are given two redox reactions and we need to identify the oxidizing agent and the reducing agent in each redox reaction. We know the oxidizing agent means the substance itself gets reduced and force to other substance to oxidation and reducing agent is the substance which oxidize itself and force to other for reduction so, we need to look for which one gets oxidized and which one gets reduced.

i)8H⁺(aq) + [Cr₂O₇]^2-(aq) + 3SO₃^2-(aq) =>2Cr³⁺(aq) + 3SO₄^2-(aq) + 4H₂O(l)

In this one there is S in the SO₃^2- is +4 and it is converted to +6 in the SO₄^2-, so SO₃^2- gets oxidized, so it is acts as reducing agent. There is Cr in the [Cr₂O₇]^2- is +6 and it is converted to 3+, so [Cr₂O₇]^2- gets reduced, means it is acts as oxidizing agent.

(ii) NO₃^-(aq) + 4Zn(s) + 7OH^-(aq) + 6H₂O(l) => 4Zn(OH)₄^2-(aq) + NH₃(aq)

In this one there is Zn in the reactant side is in free state 0 and it is converted to +2 in the Zn(OH)₄^2- , so Zn(s) gets oxidized, so it is acts as reducing agent. There is N in the NO₃^- is +5 and it is converted to 3-, so it NO₃^- gets reduced, means it is acts as oxidizing agent.