Balancing Equations with Half-Reaction and Oxidation Number Methods

In summary, Yoshi6 is having trouble with balancing equations using either the half-reaction or the oxidation number method. He asks for help from the community and receives it from Read. He is then able to solve the equation using TexAide.
  • #1
yoshi6
63
0
Hello, I am new at this, I am not exactly sure how this works. I have a question where I have to balance two equations using either the half-reaction or the oxidation number method... they are as follows.:

1. Co + MnO4- + H+ = Co+2 + Mn+2 + H2O

2. Cu + HNO3 = Cu(NO3)2 + NO + H2O

I am having trouble with the first one. I think I get the second one okay; I am not sure how to type the whole process so here is my final answer for 2;

3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

Thankyou
 
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  • #3
thanks...umm...is my second one right? If you don't mind checking?
 
  • #4
Your answer is incomplete. You were asked to balance the equations using one of the two methods. That implies that you are to show your work not just the final answer. The equation is mass balanced as you have shown it.
 
  • #5
okay I hope this makes sense:

0 +5 +2 +2
Cu + HNO3 ---> Cu(NO3)2 + NO + H20

0 -2e +2
Cu --------> Cu(-2e) x 3 = -6e

+5 +3e +2
N-----------> N(+3e) x 2 = +6e

3Cu + 2HNO3 ---> 3Cu(NO3)2 + 2NO + H2O (incomplete)
3Cu + 8HNO3-----> 3Cu(NO3)2 + 2NO + H2O (incomplete)

3Cu + 8HNO3-----> 3Cu(NO3)2 + 2NO + 4H2O
 
  • #6
it didn't work out as well as I hoped. I find it difficult to type these kinds of equations. Sorry.
 
  • #7
yoshi6 wants help with:
1. Co + MnO4- + H+ = Co+2 + Mn+2 + H2O

For the "typing" of the symbolisms, try using TexAide; but for the analysis to balance the reaction equation, use this information:
Cobalt changed its state from 0 to +2.
The manganate changed to become a +2 Manganese (not sure if can call "manganic" ion);
Each oxygen carries -2, and there are 4 oxygens on permanganate, but the permanganate net charge is -1 (as you have indicated). So you want to find the state for the permanganate:
-1 = x + 4(-2)
x = +7

So, next, balance the change in electrons for each half reaction between Co and MnO4-;
Co looses 2 electrons; Mn goes from +7 to +2, so it gained 5 electrons. You must balance the changes in electrons.
For the Co half, multiply by 5; for the Mn half, multiply by 2.

Maybe someone else could explain the balancing of oxygens and water and either OH- or H+, or maybe you can balance that part yourself.
 
  • #9
[tex]Co + Mn^{7+}O_4^- + H^+ \rightarrow Co^{2+} + Mn^{2+} + H_2O[/tex]

First we have to find which species is being reduced, and which is being oxidized. through this we'll know which species is gaining/losing electrons.

[tex]Co^{0} + Mn^{+7} \rightarrow Co^{+2} + Mn^{+2}[/tex]

Co loses 2 electrons, Mn gains 5 electrons. balance Co and Mn so that they have equal amount of electrons, but that won't be your final step.
 
Last edited:
  • #10
Thank you I get it now
 
  • #11
where is your new question?
 
  • #12
oh it is under math and science tutorials...I thought that was where I was supposed to put it? Maybe not...sorry
 

1. What is the purpose of balancing equations using the half-reaction method?

The half-reaction method is used to balance redox equations, where electrons are transferred between reactants. It allows for the balancing of both mass and charge on each side of the equation, ensuring that the chemical equation is accurate and follows the law of conservation of mass.

2. How do I identify which elements are being oxidized and reduced in a redox equation?

In a redox reaction, the element that loses electrons is being oxidized, while the element that gains electrons is being reduced. This can be identified by comparing the oxidation states of each element before and after the reaction. The element with a higher oxidation state after the reaction has been oxidized, while the element with a lower oxidation state after the reaction has been reduced.

3. What is the oxidation number method of balancing equations?

The oxidation number method is another approach to balancing redox equations. It involves assigning oxidation numbers to each element in the reaction and then using these numbers to determine how many electrons are lost or gained by each element. The goal is to balance the overall charge on each side of the equation, while also balancing the number of each element present.

4. Can the half-reaction and oxidation number methods be used together?

Yes, the two methods can be used together to balance more complex redox equations. The oxidation number method can be used to determine the overall change in oxidation states for each element, while the half-reaction method can be used to balance the transfer of electrons between reactants.

5. Are there any tips for balancing equations using these methods?

One tip is to start by balancing the elements that appear in only one reactant and one product. Then, balance the elements that appear in both the reactants and products. It can also be helpful to check the charge on each side of the equation and adjust the coefficients accordingly. Additionally, it is important to double-check that all elements and charges are balanced at the end of the process.

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