[thegunners2004]

Can someone please check my final answer to see if it is correct. If it is not, please provide me with the correct answer.

I was asked to balance this redox reaction.

Cr2O7-2 (aq) + Pb+2 (aq)-------> Cr+3 (aq) +Pb4+ (aq)


My answer is

6Cr2O7-2 (aq) +22H+(aq) + 2e + 3Pb2------------>4Cr3+(aq) + 2H2O + 3Pb4+


Thanks

[Yggdrasil]

Electrons should not be products or reactants in the final redox reaction.  Plus, the number of chromium atoms (among others) is unbalanced.  What are your two half reactions?

[thegunners2004]

my two half reactions are:

Cr2O7 -2 ------>Cr+3

Pb+3-----------> Pb+4

[Yggdrasil]

The first step would be to balance them.  If the reaction takes place in acidic medium, you can add water and H⁺ to balance out the oxygens.  If the reaction takes place in a basic medium, use water and OH^- to balance the oxygens.

Next, multiply the reactions by constant numbers so that the number of electrons produced by the oxidation reaction is equal to the number of electrons used in the reduction reaction.

Finally, add the two reactions together.  The electrons should cancel out so that the final redox reaction does not have electrons as reactants or products.

[edit: fixed a typo]

[thegunners2004]

Thanks.It helps to know that electrons can't be present in the final equation...

[AWK]

6Cr2O7-2 (aq) +22H+(aq) + 2e + 3Pb2------------>4Cr3+(aq) + 2H2O + 3Pb4+

You have 6x7=42 oxygen atoms on the left and only 2 on the right, and so on. Your equation is bad balanced.