(adsbygoogle = window.adsbygoogle || []).push({}); 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??

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# Homework Help: Electrochemistry help, Need steps

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