Balancing Equations with Half-Reaction and Oxidation Number Methods

[yoshi6]

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

 

[chemisttree]

Read this thread for an explanation of this type of problem.

https://www.physicsforums.com/showthread.php?t=174318

 

[yoshi6]

thanks...umm...is my second one right? If you don't mind checking?

 

[chemisttree]

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.

 

[yoshi6]

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

 

[yoshi6]

it didn't work out as well as I hoped. I find it difficult to type these kinds of equations. Sorry.

 

[symbolipoint]

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.

 

[Gokul43201]

...+2 Manganese (not sure if can call "manganic" ion)...

"manganous"

For writing equations: https://www.physicsforums.com/showthread.php?t=9021

 

[rocomath]

$$Co + Mn^{7+}O_4^- + H^+ \rightarrow Co^{2+} + Mn^{2+} + H_2O$$

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.

$$Co^{0} + Mn^{+7} \rightarrow Co^{+2} + Mn^{+2}$$

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.

 

[yoshi6]

Thank you I get it now

 

[rocomath]

where is your new question?

 

[yoshi6]

oh it is under math and science tutorials...I thought that was where I was supposed to put it? Maybe not...sorry