Balancing redox equation

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
 
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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 transfered 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.
 
Thank You so much:smile: !! I wish my book had explained it in steps like that.
 
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
 
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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.
 
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
 
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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.
 
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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).
 
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For this one

P4 > PH3 + H2PO2{-1}

Would I do this:

P4 > PH3
P4 > H2PO2{-1}
 
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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.
 

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