Oxidizing and Reducing agent

Question

3H2C2O4 + 2HNO32NO + 6CO2+ 4 H2O


In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

name of the element oxidized: name of the element reduced:

formula of the oxidizing agent: formula of the reducing agent:

Answer 1

First off, you are missing the arrow in the above reaction, it should be as follows: 

3 H2C2O4 + 2 HNO3 -> 2 NO + 6 CO2+ 4 H2O

 

To answer this question, I would first draw out the molecules.  The oxidation states of the C and N can be found by counting the number of bonds they have with oxygen. This should give you the following oxidation states for C and N:

      +3               +5            +2         +4  

3 H2C2O4 + 2 HNO3 -> 2 NO + 6 CO2+ 4 H2O

 

The C increased from +3 to +4 in oxidation state, so it was the element oxidized.

The N decreased from +5 to +2, so it was reduced.

HNO3 oxidized H2C2O4, so HNO3 is the oxidizing element.

H2C2O4 reduced HNO3, so H2C2O4 is the reducing agent.