Write an overall reaction oxidizing lactate to bicarbonate while reducing MnO2(s) to Mn++?

Question

Assume reaction driven by bacterial culture in an aqueous solution.

Answer 1

Lactate is CH3-CHOH-CO2-

This procedure ALWAYS works: "Start at the top and work down"; but keeping count is tricky, so check my arithmetic.

Balance atoms other than O and H (C in this case)

C3H5O3 - = 3 HCO3 -

Balance O using water

C3H5O3 - + 6 H2O = 3 HCO3 -

Balance H using H+

C3H5O2- + 6 H2O = 3 HCO3- + 14 H+

Balance charge using electrons:

C3H5O2- + 6 H2O = 3 HCO3- + 14 H+ + 12 e- (A)

Likewise, for MnO2,

MnO2 + 4H+ + 2e- = Mn2+ + 2 H2O (B)

Since electrons lost = electrons gained, multiply (in this case) (B) by 6, add up and cancel out anything that's there on both sides (in this case, you will have 14 H+ that you can cancel, as well as some water).

Check that all atoms balance, as well as charge, and you will knowthat you've got it right.

That's it!