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.

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.