PLEASE HELP WITH THESE CHEMISTRY QUESTIONS?!?!?

In a redox reaction, the oxidizing agent gets reduced, and the reducing agent gets oxidized.

Fe2O3 is the oxidizing agent, and CO is the reducing agent

Cl2 is the oxidizing agent, and KI is the reducing agent

See if you can figure out the 3rd one on your own

Cl2 + 2 e- --> 2 Cl- (reduction)

Cu --> Cu2+ + 2 e- (oxidation)

1.For the following reactions, identify the substance that is the oxidizing agent and the substance that is the reducing agent.

Fe2O3(s) + CO(g) --> Fe(s) + CO2(g)

In Fe2O3, the charge of iron is +3. In Fe, the charge of iron is 0. During this reaction, the charge of iron decreased by 3.

Fe+ 3 + 3 e- = Fe

During this reaction, the iron gained 3 electrons. A gain of electrons means the iron is reduced. This means the Fe2O3 is the oxidizing agent. And carbon monoxide is the reducing agent.

Cl2 + 2 KI = 2KCl + I2

In Cl2, the oxidation number of chlorine is 0. In KCl, the oxidation number is -1.

Cl2 + e -1 = Cl-1

During this reaction, reaction, the Cl2 lost one electron. This means Cl2 is reduced and iodine is oxidized. So chlorine is the oxidizing agent, and iodine is the reducing agent.

C2H4 + H2 ----> C2H6

To determine the oxidation number of carbon and hydrogen, we need to look at a table that has the electronegativity of the elements. The electronegativity of carbon is 2.5. The electronegativity of hydrogen is 2.1. The element with the highest electronegativity has a negative charge. This means carbon has a negative oxidation number in each of the compounds. In C2H4, the oxidation number of carbon is -2. In C2H6, the oxidation of carbon is -3. This means the oxidation number of carbon decreased. This means

C-2 + x = C-3

C-2+ e- = C-3

Gain of an electron means carbon is reduced, and hydrogen is oxidized. So C2H4 is the oxidizing agent, and H2 is the reducing agent.

2.Separate this redox reaction into its component half reactions.

Cl2+2Cu ---> 2CuCl

Oxidation half reaction:?

Reduction half reaction?

#1: Cl2 → 2 Cl-1 + 2 e-

#2: 2 Cu + 2 e- → 2 Cu

#1 is the reduction half reaction. #2 is the oxidation half reaction.

I hope this is helpful for you.