Find the oxidizing agent and the reducing agent. 2na(aq)+2h2o(l)→2naoh(aq)+h2(g) c(s)+o2(g)→co2(g) 2mno−4(aq)+5so2(g)+2h2o(l)→2mn2+(aq )+5so2−4(aq)+4h+(aq)

Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.

(1)  2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent. 

The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.

(2)  C(s) + O2(g) --> CO2(g)

The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent. 

The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.

(3)  2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺

The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.

The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent. 

The correct answer is c

No is the answer to the question