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Balance the following redox equation, using half-reactions. Assume that the reaction occurs in an aqueous solution.

Cr2O72– + NO → Cr3+ + NO3–

User Neil Benn
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Final answer:

To balance the redox equation Cr2O72– + NO → Cr3+ + NO3–, one must write the separate oxidation and reduction half-reactions, balance atoms and charges in each, ensure electrons are equal and opposite, and then combine and simplify the equation.

Step-by-step explanation:

Steps to Balance the Redox Equation

To answer the question on how to balance the redox equation Cr2O72– + NO → Cr3+ + NO3– using half-reactions, we must apply the steps for balancing redox reactions in an aqueous solution. First, we write the oxidation and reduction half-reactions. In this case, Cr2O72– is reduced to Cr3+ and NO is oxidized to NO3–.

For the reduction half-reaction (Cr2O72– to Cr3+), we balance the oxygen by adding water molecules and the hydrogen by adding H+ ions (in acidic solution), then balance the charge by adding electrons. For the oxidation half-reaction (NO to NO3–), we balance the oxygen by adding water, the hydrogen by adding H+ ions, and finally, the charge by adding or removing electrons.

This process requires the balancing of atoms and charges in each half-reaction separately, and once that is done, we combine the two half-reactions and ensure that the electrons lost in oxidation equal the electrons gained in reduction. Finally, we simplify the equation by canceling out species that appear on both sides.

User Trax
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First determine the formal oxidation numbers:
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons:
Oxidation half reaction:
NO(aq) + 2 H2O(l) ---> (NO3)-(aq) + 4 H+(aq) + 3 e-
Reduction half reaction:
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l)
Now balance the number of electrons on both sides and add them together:
2 NO(aq) + 4 H2O(l) ---> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l) --------------------------------------... 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) ---> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l)
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly:
-2 + 6 = -2 + 2(+3) +4 = +4
User Umber
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