Reduction oxidation equations?

Question

When you balance the redox equation:

C4H10(l) + Cr2O72-(aq) + H+(aq) H6C4O4(s) + Cr3+(aq) + H2O(l)

the oxidizing agent is:

Answer 1

C4H10  (C4 totals -10 charge) -->   H6C4O4 (C4 totals +2 charge) & 12 e- lost

1 (Cr2O7)-2 (2 Cr's total +12)  & 6 e- taken--> 2 Cr+3  (total +6)

to balance the electrons taken & lost, we double the (Cr2O7)-2:

C4H10  -->   H6C4O4  & 12 e- lost

2 (Cr2O7)-2   &  12 e- taken--> 4 Cr+3

then we combine them:

C4H10  &  2 (Cr2O7)-2   -->   H6C4O4  & 4 Cr+3

now we balance the charges with the H+'s,, at present the left totals -4, while the right totals +12,... we need 16 H+ 's on the left to balance the charges to +12 & +12:

C4H10  &  2 (Cr2O7)-2   &  16H+  -->   H6C4O4  & 4 Cr+3

now we we balance the H's with the neutral water:

C4H10  &  2 (Cr2O7)-2   &  16H+  -->   H6C4O4  & 4 Cr+3  &   10 H2O

every molecule has its # of moles,

as the check on the work,

the oxygens should already be balanced:

14 are on the left, & 14 are on the right, it is done, your answer is:

C4H10  &  2 (Cr2O7)-2   &  16H+  -->   H6C4O4  & 4 Cr+3  &   10 H2O

the method is:

1) balance the electons taken & lost

2) balance the charges

3)balance the misc atoms

4)saving oxygen as the check on your work

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almost forgot your second question,...

the oxidizing agent is the taker of electrons, Cr2O72-(aq)