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Question

Identify the oxidizing and reducing agents in each of these redox reactions

1. H2S + Cl2 --> 2HCl + s

2. N2 + H2 -->NH3

3. Na + I2 --> 2NaI

and i need help with these questions

4. What are the three basic types of compounds that are used as antacids?

Balance each half-reaction for atoms and charge

1. Na --> Na+

2. Sn4+ --> Sn2+

3. Fe2+ --> Fe3+

4. Cl2 --> Cl-

Answer 1

Oxidation is the loss of electrons and reduction is gaining electrons. To determine which species is being oxidized and which species is being reduced you need to assign oxidation numbers first. Read more about assigning oxidation numbers in your text book. It should take about 15 minutes and it will clear up some of your confusion.

1. H2S + Cl2 ---> 2HCl + S

first I'm going to determine oxidation numbers

reactants:

2 + S = 0

S= -2

Cl= 0

products:

2 + 2Cl = 0

2Cl = -2

Cl= -1

S=0

So going from reactants to products sulfur goes from -2 to 0, which corresponds to a loss of electrons, oxidation. Chlorine does from 0 to -1, which corresponds to gaining electrons, reduction. So the sulfur is the reducing agent and chlorine is the oxidizing agent.

2. N2 + H2 --> NH3

reactants:

N = 0

H = 0

products:

N+3=0

N=-3

H=3

So going from reactant to products, nitrogen goes from 0 to -3, which corresponds to gain of electrons, reduction, and H goes from 0 to 3, oxidation a loss of electrons. So the oxidizing agent is N and the reducing agent is H.

3. Na + I2 --> 2NaI

reactants:

Na = 0

I2 = 0

products:

2 + 2I = 0

2I=-2

I = -1

Going from reactants to products iodine goes from 0 to -1, which a gain of electrons, which corresponds to reduction. On the other hand, sodium went from 0 to 2, a loss of electrons, which is oxidation. So iodine is the oxidizing agent and sodium is the reducing agent.

Half-reactions:

1. Na --> Na+

Na goes from 0 to +1, this is oxidation so the electrons are a product.

Na --> Na+ + 1e-

2. Sn4+ --> Sn2+

Sn goes from 4+ to 2+, this is reduction so the electrons are reactants

Sn4+ + 2e- --> Sn2+

3. Fe2+ --> Fe3+

Fe goes from 2+ to 3+, this is oxidation so the electrons are products

Fe2+ --> Fe3+ + 1e-

4. Cl2 --> Cl-

Cl goes from 0 to -1, this is reduction so the electrons are reactants

Cl2 + 1e- --> Cl-

These are half-reactions of the very basic kind. Just wait until they want you to balance half-reactions in acidic or basic solution. I don't think its hard. Its actually kind of fun. In acidic solution you balance oxygen by adding water, and then you balance hydrogen by adding H+. In basic solution solution you just add equal amounts of OH- for all H+ to give water.

I'm not sure about the antacids question, sorry. You can probably find the answer in your book in the margin somewhere in one of the little passages that no one ever reads.

Hey, I think I've seen you in the LGBT section. ;)