Balancing Redox Equations: Step-by-Step Guide for Beginners

  • Thread starter Shay10825
  • Start date
  • Tags
    Redox
In summary, a redox equation is a chemical equation that shows the transfer of electrons between two or more chemical species. It is important to balance redox equations in order to follow the law of conservation of mass and charge. The steps for balancing a redox equation include identifying the oxidizing and reducing agents, writing and balancing the half-reactions, and combining them to cancel out any common terms. A redox equation is considered balanced when the number of atoms and charges are equal on both sides. Some common methods for balancing redox equations include the half-reaction method, the oxidation number method, and the ion-electron method.
  • #1
Shay10825
338
0
Hello. I have a question on balancing redox equations. I’ve tried to follow the steps in my book but I don’t understand how to do it. Can someone please walk me through one.

H2S(g) + NO3{-1}(aq) > NO(g) + S(s) (acidic solution)

Now the book says I have to add electrons but how?

Thanks
 
Physics news on Phys.org
  • #2
1. Get the half-reactions:

H2S --> S
NO3- --> NO

2. Balance any atoms that are not O or H:

H2S --> S
NO3- --> NO

3. Balance O atoms using H2O:

H2S --> S
NO3- --> NO + 2H2O

4. Balance H atoms using H+:

H2S --> S + 2H+
4H+ + NO3- --> NO + 2H2O

5. Balance charge with electrons:

H2S --> S + 2H+ + 2e-
4H+ + 3e- + NO3- --> NO + 2H2O

6. Multiply equations so that you have the same amount of electrons being transferred in each reaction:

3[H2S --> S + 2H+ + 2e-] = 3H2S --> 3S + 6H+ + 6e-
2[4H+ + 3e- + NO3- --> NO + 2H2O] = 8H+ + 6e- + 2NO3 --> 2NO + 4H2O

7. Add up equations:

3H2S --> 3S + 6H+ + 6e-
8H+ + 6e- + 2NO3 --> 2NO + 4H2O
------------------------------------
3H2S + 8H+ + 6e- + 2NO3- --> 3S + 6H+ + 6e- + 2NO + 4H2O

8. Cancel out species that occur on opposite sides:

3H2S + 2H+ + 2NO3- --> 3S + 2NO + 4H2O

9. Check for errors. Make sure atoms are balanced and the overall charge on each side is balanced.
10. This is your answer because it is in acidic solution. If it is was in basic solution, you would have to neutralize the hydrogen atoms with hydroxide ions.
 
  • #3
Thank You so much:smile: ! I wish my book had explained it in steps like that.
 
  • #4
Cesium said:
10. This is your answer because it is in acidic solution. If it is was in basic solution, you would have to neutralize the hydrogen atoms with hydroxide ions.

What would I do if it was in a basic solution. For example:

Cr{+3}(aq) + MnO2(s) > Mn{+2}(aq) + CrO4{-2}(aq) in a basic solution
 
  • #5
Ok well let's just pretend the first problem you posted was done in basic solution. For acidic solution, we had:

3H2S + 2H+ + 2NO3- --> 3S + 2NO + 4H2O

So we have 2H+ that we need to neutralize with 2OH-, giving 2H2O on the left side and 2OH- on the right side:

3H2S + 2H2O + 2NO3- --> 2S + 2NO + 4H2O + 2OH-

Now we can cancel out the 2H2O on the left with the 4H2O on the right:

3H2S + 2NO3- --> 2S + 2NO + 2H2O + 2OH-

This reaction could never occur in basic solution, but it's just an example.
 
  • #6
One last question. I was doing this one and I got stuck:

IO3{-1}(aq) + I{-1}(aq) > I2(s)

IO3{-1} + I{-1} > I2
6H{+1} + IO3{-1} + I{-1} > I2 + 3H2O
4e- + 6H{+1} + IO3{-1} + I{-1} > I2 + 3H2O

But now what do I do because there is not another e on the other side.

Thanks
 
  • #7
This one is a bit tricky. Like always, first split it up into two distinct half-reactions:

IO3- --> I2
I- --> I2

and then balance the I atoms:

2IO3- --> I2
2I- --> I2

Continue from here and I think you should be able to get it. Tell me if you have problems.
 
Last edited:
  • #8
Yeah I can finish that one. Would I do the same thing for one like this:

P4 > PH3 + H2PO2{-1}

And how do I know what is the reducing agent and oxidizing agent in the final balanced equation (you can use the 1st one for an example since it is already balanced).
 
Last edited:
  • #9
For this one

P4 > PH3 + H2PO2{-1}

Would I do this:

P4 > PH3
P4 > H2PO2{-1}
 
  • #10
Shay10825 said:
Would I do this:

P4 > PH3
P4 > H2PO2{-1}
Yes.

IO3- --> I2
IO3-: I has an oxidation state of +5.
I2: I has an oxidation state of 0.

I is moving towards a more negative state, so it is reduced. Therefore, IO3- is the oxidizing agent.

I- --> I2
I-: I has an oxidation state of -1
I2: I has an oxidation state of 0

I is moving towards a more positive state, so it is oxidized. Therefore, I- is the reducing agent.

Same procedure applies to your P4 one.
 

1. What is a redox equation?

A redox equation is a chemical equation that shows the transfer of electrons between two or more chemical species. It involves the oxidation and reduction of atoms or ions, resulting in a change in their oxidation states.

2. Why is it important to balance redox equations?

Balancing redox equations is important because it ensures that the number of atoms of each element and the total charge are equal on both sides of the equation. This follows the law of conservation of mass and charge, which states that matter cannot be created or destroyed.

3. What are the steps for balancing a redox equation?

The steps for balancing a redox equation are: (1) Identify the oxidizing and reducing agents, (2) Write the half-reactions for the oxidation and reduction processes, (3) Balance the atoms in each half-reaction, (4) Balance the charges by adding electrons, (5) Multiply the half-reactions by appropriate coefficients to balance the number of electrons, and (6) Combine the half-reactions and cancel out any common terms.

4. How do you know if a redox equation is balanced?

A redox equation is considered balanced if the number of atoms of each element and the total charge are equal on both sides of the equation. This can be checked by counting the number of atoms and charges on each side of the equation and ensuring that they are equal.

5. What are some common methods for balancing redox equations?

Some common methods for balancing redox equations include the half-reaction method, the oxidation number method, and the ion-electron method. Each method follows a set of steps to balance the equation and may be more suitable for certain types of redox reactions.

Similar threads

  • Biology and Chemistry Homework Help
Replies
8
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
  • Biology and Chemistry Homework Help
Replies
1
Views
3K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Biology and Chemistry Homework Help
Replies
18
Views
4K
  • General Math
Replies
3
Views
4K
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
  • Biology and Chemistry Homework Help
Replies
4
Views
5K
Back
Top