Balance the following redox equation, using half-reactions. Assume that the reaction occurs in an aqueous solution.
Cr2O72– + NO → Cr3+ + NO3–

Question

Balance the following redox equation, using half-reactions. Assume that the reaction occurs in an aqueous solution.
Cr2O72– + NO → Cr3+ + NO3–

anonymous · Answer

It is fairly a simple equation after having balanced it.
On the left hand side you have 8 oxygen while on the right hand side you have 3 oxygen, so add 5 water molecule to right hand side to balance the number of oxygen molecules.
And multiply Cr3+ by 2 to balance the Cr atoms. Now you have 
you have 
(Cr207)^-2 +  NO-->2Cr3+  +  (NO3)^-   +   5H2O

Now that you have balanced the equation, you can then balance the charges.
On the right hand side you have a total of +5, while on the left hand side you have -2. Add 10 H+ on the left hand side to balance the hydrogen atoms, while 7e on the right hand side to balance the charges shown below:
(Cr207)^-2 +NO + 10H+-->2Cr3+ +(NO3)^- + 5H2O +7e