Can someone me with these 2 questions?

for each of the following redox reactions, determine which element is being reduced and which is being oxidized.

2 mno4−(aq) + 5 c2o42−(aq) + 16 h3o+(aq) → 2 mn2+(aq) + 10 co2(g) + 24 h2o(l)

3 cu(s) + 8 hno3(aq) → 3 cu(no3)2(aq) + 2 no(aq) + 4 h2o(l)

1.

2MnO4−(aq) + 5C2O42−(aq) + 16 H3O+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 24 H2O(l)

Element being reduced -MnElement being oxidized - CExplanation:Redox reactions are reactions that involves both oxidation and reduction.Reduction involves the gain of electrons by an atom while oxidation is the loss of electrons.A reduction may also refer to a decrease in the oxidation number of an element while oxidation is an increase in the oxidation number of an element in a reaction.In the Equation:

2MnO4−(aq) + 5C2O42−(aq) + 16 H3O+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 24 H2O(l)

The oxidation number of Mn on the left side of the equation is +7 and on the right side of the equation is +2, therefore Mn has undergone reduction.On the other hand, the oxidation number of Carbon (C) on the left side of the equation is +3 and on the right side of the equation is +4, therefore Carbon(C) has undergone oxidation.2. Answer

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(aq) + 4 H2O(l)

Element being reduced - NElement being oxidized - CuExplanation In the Equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(aq) + 4 H2O(l)

The oxidation number of Cu on the left side is 0 and on the right side is +2, therefore, copper has undergone oxidation.The oxidation number of Nitrogen on the left side is +5 and on the right side is +2, therefore, nitrogen has undergone reduction.