Question

Balance the redox equation, using half reactions. Assume reaction occurs in aqueous solution.

Cr2O7^2- + NO = Cr ^3+ + NO^3-

Don't understand how to do it.

Answer 1

Half reactions show whether something is oxidized (loss of electrons) or reduced (gain of electrons. The final half reactions should be this:

14H^+ 6e^- + Cr2O7^2- -> 2Cr^3+ + 7H2O
NO + 3e^- -> NO^3-

The steps:
First, separate the reactions and balance the atoms other than oxygen or hydrogen.
Cr2O7^2- -> 2 Cr^3+
NO -> NO^3-

Next, balance the amount of oxygen by adding water on one side if necessary
Cr2O7^2- -> 2 Cr^3+ + 7H2O
NO -> NO^3-

Now balance the hydrogens by adding protons
14H+ + Cr2O7^2- -> 2Cr^3+ + 7H2O
NO -> NO^3-

Finally, make sure the charges are equal on both sides. Add electrons to one side as needed. For the Cr reaction, the reactants have an overall charge of 12+, products 6+. You need 6 electrons on the reactant side to balance this out. For the NO reaction, you have a charge of 0 on one side, -3 on the other. Add 3 electrons to the reactant side.

Finally, you get

14H^+ 6e^- + Cr2O7^2- -> 2Cr^3+ + 7H2O
NO + 3e^- -> NO^3-

You can combine them back into one equation if needed. Hope this helped. :)

Answer 2

reduction half reaction should be

2H2O + NO ==== NO3^3- + e + 4H^+