Complete the reduction half-reaction by identifying the correct coefficient for the electr… — JAMB UTME

Complete the reduction half-reaction by identifying the correct coefficient for the electrons needed to balance the charge.

Answer:In the balanced reduction half-reaction in acidic medium:

\text{MnO}_4^- + 8\text{H}^+ + 【5】e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

, the missing coefficient for the electrons is 【5】.

The total charge on the left side of the equation is

+7

(from one

\text{MnO}_4^-

ion carrying

-1

and eight

\text{H}^+

ions carrying

+8

). The total charge on the right side is

+2

(from one

\text{Mn}^{2+}

ion). To make both sides equal in charge, 5 electrons (each having a

-1

charge) must be added to the reactant side.

Calculate the total ionic charge on the reactant side before adding electrons.

Net reactant charge =

(-1) + 8(+1) = +7

One permanganate ion contributes a charge of

-1

and eight hydrogen ions contribute

+8

Calculate the total ionic charge on the product side.

Net product charge =

+2

One manganese(II) ion contributes

+2

and four water molecules are neutral (

0

Determine the number of electrons required to balance the overall charge.

+7 + 5(-1) = +2

, so 5 electrons are required.

Electrons carry a

-1

charge, so adding

5e^-

to the reactant side lowers its charge from

+7

+2

to match the product side.

Balancing charge in half-reactions by adding electrons

Question