# Homework Help: Electrochemistry help, Need steps

1. Jun 19, 2011

### Nelo

1. The problem statement, all variables and given/known data

Several problems.

1) Balance these oxidation-reduction equations in both acidic and basic conditions. In each equation,
underline the oxidizing agent. (I have highlighted them. Remember, the oxidizing agent is the species that
is reduced and gains electrons.)

a)Cr2o7-+ C2H5OH → Cr3+ + CO
b) HNO3 + P -> No + h3po4.
c) C2h4 + MNO4- -> Mn2+ + CO2

2. Relevant equations
No equations needed

3. The attempt at a solution

I just seem to not understand what the hell is going on. We talked about this breifly but it will be on the exam ( on tuesday) Here is what I got from what she taught...

a) balance all elements except hydrogen and oxygen
b) Balance oxygen using H2o
c) Balance hydrogen using H+
d) balance the charge of rctnt side with charge of product side by adding electrons to either reactant or product
e) Multiply 1 or both balanced half reactions by integer to equalize number of electrons transfered in the two half reactions.

So..#1 in acidic :: Cr2O7-+ C2H5OH → Cr3+ + CO

Cr2O7 ---> Cr3+ First i found charge of Cr in Cr2O7 (which is 6) . since difference of 6 and 3 is 3. and it went from +6 to +3 it gained 3 electrons and was reduced. so the half rctn would be

14H+ + Cr2O7 + 3e ----> Cr3+ + 7H2O

Second half reaction.. : H2C2O4 --> 2 CO2 + 2 H+ + 2 e- ( I dont even understand why thats 2 electrons...?)

Thats basically all i understand .. and thats just the beginning part. Can anyone help me to get to the right steps of this??

2. Jun 19, 2011

### Staff: Mentor

This is not balanced.

2 electrons to balance charge. Properly balanced reaction (or half reaction) must be balanced in terms of atoms AND charge. There is no charge on the left (total of zero), so you need zero on the right. After balancing atoms you have 2H+ there - that means charge of +2, to neutralize it you need to add two electrons.

When balancing half reactions you don't use oxidation numbers, they aren't needed. First balance atoms, then balance charge adding electrons on one of the sides.

See:

balancing equations with half reactions method

balancing equations with oxidation numbers method

3. Jun 19, 2011

### Nelo

Are you always balancing the oxidation one by multiplying it by set amount of value (# of electrons ) . ?

4. Jun 19, 2011

### Nelo

Also, to completely balance that in terms of electrons you would multiply the dicromate half reaction by 2 , and the co2 half reaction by 3 ? giving 6 electrons? thus cancelling

5. Jun 19, 2011

### Nelo

I still dont understand it.

Cr2O7 2- + C2H5OH -> Cr3+ + CO2.

1) 14H+ + Cr2O7-2 ----> 2Cr 3+ + 7H2O .
the only thing left is the electrons

The overall charge on the left side is (14) (-2) = 12. overall charge on right is (3) (0) *unless the coefficnat multiplies with the 3+ giving 6 on the right, thus needing 6 electrons to be stable. (resulting in multplying the other half rctn by 3)

6. Jun 19, 2011

### Nelo

I did a seperate problem. Seem to have gotten the right answer. Please tell me if you notice something off.

Problem : As2O3 + NO3- ---> H3AsO4 + NO

What i did : 4H+ + No3- ---> NO + 2H2o +3e

2nd half : 5H2o + As2o3 + 4e -----> 2H3AsO4 + 4H+

Since we need electrons to balance, multiply top half by 4, and bottom half by 3, then combine and simplify.

Giving : 3 As2O3 + 4 NO3- + 7H2o +4H+ ----> 6H3AsO4 + 4NO . I'm going to try doing basic solution now.

7. Jun 20, 2011

### Staff: Mentor

Charge of 14H+ means 14*+1 - you did that correctly. Why do you have any doubts what is the charge of 2Cr3+? 2*+3 it is, we ara talking about total charge, not a single ion cherge.

What is charge on the left? What is charge on the right? Are they identical?

8. Jun 20, 2011

### Nelo

C2H4 + MnO4-  Mn2+ + CO2

How do you solve this one?

I get to 8H+ + MnO4- ----> Mn 2+ + 4H2O
(7 charge) (2 charge) needs 5 electrons.

2nd half being 4H2O + C2H4 ----> 2CO2 + 12H+
needs 12 electrons.

This doesnt make sense.

9. Jun 21, 2011

### Staff: Mentor

Yes, it needs 12 electrons.