Identify the oxidizing agent in the reaction:

8H^+ (aq) + 6Cl^- (aq) + Sn(s) + 4NO3^- (aq) ---> SnCl6^2- (aq) + 4NO2(g)+4H2O(l)

oxidizing agent:______ reducing agent:______

2MnO4^- (aq) + 10 Cl^- (aq) +16 H^+ (aq) --> 5 Cl2(g) + 2 Mn^2+(aq) + 8 H2O(l)
oxidizing agent:_______ reducing agent:_____

i'm really confused on how to do these problems. Please help, thank you

Start by identifying the element oxidized and the element reduced. The element oxidized is contained in the reducing agent, the substance that causes something else to become reduced. The element reduced is in the oxidizing agent, the substance that causes something else to become oxidized.

8H^+ (aq) + 6Cl^- (aq) + Sn(s) + 4NO3^- (aq) ---> SnCl6^2- (aq) + 4NO2(g)+4H2O(l)

Tin is oxidized, and is therefore, the reducing agent. Nitrogen is reduced and therefore, nitrate ion is the oxidizing agent.

In the second reaction Mn is reduced, and MnO4- is the oxidizing agent, while chloride ion is oxidized, and is therefore, the reducing agent.