GChem Questions about Redox

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Acheiver Test 3 Question 48
Balance Redox and get sum of coeff for reactants and products:
Al + NO3- + H+ -----> N2O4 + H2O + Al3+

In their solution one of the half reaction is
2e- + 4H+ + 2NO3- ------> N2O4 + 2H2O
My question is, why are two e- needed to balance it? The oxid numbers go from +5 to +4, isnt that just 1 e-.

Destroyer Question 70:
Consider the reacyion of 2.5 M dichromate:
(Cr2O7)-2 ----> Cr+3 (Acidic Medium

What would be the solution Normality if the reaction goes to completion?

The solution shows the balanced reaction to be
6e- +14H+ + (Cr2O7)-2 -------> 2Cr+3 + 7H2O

But instead of multiplying the M by 6, they multiply it by 3.

 

[ethanyoo726]

Ok so for your Destroyer Question, you need to look at the oxidation number of Cr. In Cr2O7, the oxidation number of Cr is +6. On the other side of the equation, Cr becomes +3, which means it gains 3 electrons. So you multiply 2.5 by 3 to get 7.5N.

For your Achiever Test Question, each half reaction balanced should look like this:
1) 3Al --> Al3+ + 1e-
2) 4H+ + 2NO3- + 2e- --> N2O4 + 2H2O

The reason for 2 e- is because for the equation #2, if you take out the 2e-, there is a +4 (from hydrogens, which are +1 each) and a -2 (from the nitrates, which are -1 each), so there is an overall difference of +2 on the left side. On the right side of the equation you have oxidation number of 0 for N2O4 and 0 for H2O, so an overall 0 on the right side. So in order to balance the left and right side, you need to add 2 e- to the left side of the equation to cancel out the +2. that way, you balance the equation.

So when you want to balance both equation #1 and #2, you multiply equation #1 by 2, in order to match the 2e- from the equation #2. You're overall reaction should look like this...

1) 2 [ 3Al --> Al3 + 1e-] = 6Al --> 2Al3 + 2e-
2) 4H+ + 2NO3- + 2e- --> N2O4 + 2H2O

so after cancelling, you're overall net reaction should look like...

6Al + 2NO3 + 4H+ --> N2O4 + 2Al3

i hope this helps.