need help balancing two reaction equations.

[nubbs707]

the question states: add these two equations together to make a balanced equation. the electrons should cancel out, so you will have to multiply each half reaction by a small whole number before adding the two equations together.

the reaction equation are:
Mn(s) -> Mn²⁺ + 2^-
HNO{3} + 3H⁺ + 3^- -> NO + 2H{2}O

can any one explain 1st what exactly im supposed to be determining and 2nd how to balance it out?

thanks

[sharan]

Hey you are supposed to add these two half reaction in order to make a full reaction so for example in your case you have Mangnese being oxidized and nitric acid (HNO₃) being reduced. So your purpose is to combine both of these or "add" these two reactions but you have to make sure that there are suposed to be no electrons (negative charge) on either side of the reaction. If you see in first half reaction you have 2 electrons on right side and in second you have 3 electrons on left side. Right. Now if you just add the reactions, 3 electrons on left and 2 electrons on right, even after canceling out there would still be one electron left on the left side. So you have to think how can you make same electrons on both sides. Therefore, you have to think a comoon factor of 2 and 3 (the smallest one) read the table of 2. 2, (not in 3 ) 4, (not in 3) 6 (it is in 3). So you common factor must be 6. Now to make the first half reaction electrons (2 electrons) to be 6 you hvae to multiply it by 3 since (2 x 3 =6) and for the second equation you hvae 3 electrons you must multiply it by 2 (3 x 2 =6) this way when you are adding the reaction both side electrons will cancel out. This is electron balancing but the issue is that wen you are multiplying the first equation by 3, you are not just multiplying the electrons, it is the whole equation that is being multiplied and similarly the seconf one. So it will become sumting like:

3Mn(s) -> 3Mn²⁺ + 6^-
2HNO{3} + 6H⁺ + 6^- -> 2NO + 4H₂O

Now you just have to add these like:
3Mn(s) + 2HNO{3} + 6H⁺ + 6^- -> 3Mn²⁺ + 6^- +2NO + 4H₂O

(count every element, it must be balanced on both sides)
now just cancel the similar terms on both sides, you will cancel out the electrons , then your balanced redox reaction becomes
3Mn(s) + 2HNO{3} + 6H⁺ -> 3Mn²⁺ +2NO + 4H₂O

- hope that helps.

-Sharan

[nubbs707]

thanks sharon that cleared it up perfectly!!!!!