Electron transfer in Redox equation

[ally1h]

Homework Statement

How many electrons are transferred in the following reaction?
6 Br^- (aq) + Cr2O7^2- (aq) + 14 H^+ --> 2Cr^3+ (aq) + 7H2O + 3Br2 (l)

a) 1
b) 2
c) 3
d) 6
e) 14

Homework Equations

The Attempt at a Solution

Redox reactions really confuse me, but I gave it a shot. I took a look at the number of electrons on each side of the equation and found the following:
Left side:
6Br^- = -6
Cr2O7^2- = -2
14 H^+ = +14
Total: +6

Right side:
2Cr^3+ = +6
7H2O = 0
3 Br2 = 0
Total: 6

But this is as far as I can get. Somehow I don't think the answer is D) 6, but I really don't know. Someone please help me understand this problem?

 

[symbolipoint]

Easily concentrate your attention to just one half-reaction, assuming the reaction as written is balanced; that being for the bromide. How many electron difference is it from 2 bromides to one bromine(compound, not the separate atoms)? Now, how many electrons change is this for the same half-reaction as represented in the fully written reaction (in which you start with 6 bromides instead of just 2 bromides)?

 

[symbolipoint]

In other words, from post #2, your best choice would seem to be D. (6 electrons)

 

[ally1h]

Okay, so D it is. If I understand correctly then in the half-reaction for bromide... there is 1 electron difference between 2 bromides to 1 bromine. As a whole reaction, there are 6 bromides... so 6 electrons.

What about the other half-reaction, though?
Cr2O7^2- + 14H^+ --> 2Cr^3+ +7H2O

I suppose it is the dichromate ion that confuses me.

 

[symbolipoint]

If you know your reaction as written is balanced, then you know that the chromate part of the reaction is also for 6 electrons.

Look at dicrhromate anion. Account for all the charges which give the -2 charge for this anion. Two chromiums, seven oxygens, the sum of the charges must be -2. You want to find the charge on the chromium. You know: 1 oxygen is -2, dichromate anion charge is -2. You do not yet know the charge on 1 chromium in the dichromate anion. This is what you want to find (using simple algebra).

 

[ally1h]

Ahhh... I see. Thank you for your help. I understand now. :) Redox reactions make my head spin.