Which of these oxidation half-reactions is correct? Please explain?

Question

a. Cr(NO3)3 + 4H2O --> CrO4 (-2) + 3NO3 (-3) + 8H (+) + 3e-

b. Cr(NO3)3 + 10H (+) --> CrO4 (-2) + 3N (5+) + 5H2O + 3e-

Im not sure if im dealing with the Nitrogens appropiately.

Answer 1

The best way to look at this reaction is by looking at the net ionic equation. The NO3- ions are going to be spectator ions, so you do not need to be concerned with them.

The reaction is Cr(3+) --> CrO4(2-).

The chromate ion only exists in basic solution, so at the end, you will have hydroxide ions.

We start by balancing the equation as if it was in acidic solution and later clean it up.

So, we must balance oxygen first by adding water.

Cr(3+) + 4 H2O --> CrO4(2-)

Now we balance the hydrogens by adding H(+)

Cr(3+) + 4 H2O --> 8 H(+) + CrO4(2-).

Since the reaction is taking place in basic solution, we cannot have H(+) floating around, to remove them, we add as many (OH-) to both sides of the equation.

Cr(3+) + 8 OH(-) + 4 H2O --> 8 H(+) + 8 OH (-) + CrO4(2-)

Now, H(+) and OH(-) when they encounter each other immediately form water, so the reaction becomes.

Cr(3+) + 8 OH(-) + 4 H2O --> 8 H2O + CrO4(2-)

Since we have water molecules on both sides of the arrow, four of them are going to cancel out, so we get:

Cr(3+) + 8 OH(-) --> 4 H2O + CrO4(2-)

Now we just balance the charges.

The reactants have 5 negative charges, the products just two, since the Cr(3+) increased its oxidation number to +6 in the CrO4(2-), that tells me I need 3 electrons to balance this oxidation half reaction, so:

Cr(3+) + 8 OH(-) --> CrO4(2-) + 4 H2O + 3 e(-)

We are done. Everybody is balanced by mass and charge.

Answer 2

1. You can't have (CrO4)2- and H+ together on the same side. Either OH- (or (Cr2O7)2-).

2. Why did you leave the nitrate ions in on the left hand side? Get rid of them.

2Cr3+ + 7H2O ----> (Cr2O7)2- + 14H+ + 6e-.