Multiply each equation by the number of electrons in the other equation, then combine the two equations in order to cancel the electrons out.

Question

 Multiply each equation by the number of electrons in the other equation, then combine the two equations in order to cancel the electrons out.

   (Just like if you were trying to solve for x and you had the equations 5x + 3y = 15 and 2x - 2y = 6. You would multiply the first eqn by 2, the second by 3 and then add them to cancel out y.)

Combined Steps:  (Determine the numbers that go in the blanks. If zero, record as the number zero "0".)

 _____ MnO₄ ^- + _____H⁺ + _____e^-  + _____ R-H3 + _____H₂O --> _____Mn²⁺ + _____H₂O + _____HO-R=O + _____H⁺ + _____e^- 

Answer 1

This is a REDOX reaction and can be balanced as such. (I must say that the way the question is phrased is "confusing", but I can show you what can be done to answer all it's parts & get the balanced equation). First, we will separate out the two half-reactions:

MnO₄^- →→ Mn²⁺ and RH₃ → HOR=O (which I will condense to HRO₂ for simplicity).

Now, we will balance each half-reaction by adding water molecules to the side needing O atoms, and H⁺ to the side needing H-atoms, and then add electrons (e^-) to balance the charge.

MnO₄^- + 8 H⁺ + 5 e^- → Mn²⁺ + 4 H₂O

RH₃ + 2 H₂O → HRO₂ + 6 H⁺ + 6 e^-

Now, to make the gain of electrons (reduction) equal the loss (oxidation), we need to multiply the top half-reaction by 6 and the other by 5. This now gives us:

6 MnO₄^- + 48 H⁺ + 30 e^- → 6 Mn²⁺ + 24 H₂O

5 RH₃ + 10 H₂O → 5 HRO₂ + 30 H⁺ + 30 e^-

As you see, the gain & loss of electrons are equal (& can be "cancelled" out). Likewise, we can subtract out water molecules & H⁺ ions, just as you would with algebraic equations having like terms on both sides. This results in the following:

6 MnO₄^- + 18 H⁺ + 5 RH₃ → 6 Mn²⁺ + 5 HRO₂ + 14 H₂O

The equation is now balanced in charge & mass.