How many mL of 0.125 M Cr3+ must be reacted with 12.00 mL of 0.200 M MnO4- if the redox products are CrO42- and Mn2+?

First, write and balance the redox reaction:

Cr3+ + MnO4- → Mn2+ + CrO42-

To balance this reaction, we will use the half-reaction method:

Oxidation half-reaction: Cr3+ → CrO42- + 3e-
Reduction half-reaction: MnO4- + 5e- → Mn2+

Balance the oxygen atoms by adding water molecules:

Cr3+ + 4H2O → CrO42- + 8H+ + 3e-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Balance the charges by adding electrons:

Cr3+ + 4H2O → CrO42- + 8H+ + 3e-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 3 to balance the electrons:

5Cr3+ + 20H2O → 5CrO42- + 40H+ + 15e-
3MnO4- + 24H+ + 15e- → 3Mn2+ + 12H2O

Add the two half-reactions and simplify:

5Cr3+ + 3MnO4- + 8H2O → 5CrO42- + 3Mn2+ + 16H+

Now we can use stoichiometry to solve the problem.

Moles of MnO4- = (0.200 mol/L) * (12.00 mL) * (1 L/1000 mL) = 0.00240 mol

From the balanced equation, the mole ratio of Cr3+ to MnO4- is 5:3.

Moles of Cr3+ = (0.00240 mol MnO4-) * (5 mol Cr3+/3 mol MnO4-) = 0.00400 mol

Volume of Cr3+ = (0.00400 mol) / (0.125 mol/L) = 0.0320 L = 32 mL

Therefore, 32 mL of 0.125 M Cr3+ must be reacted with 12.00 mL of 0.200 M MnO4-.