JAMB - Chemistry (2000 - No. 24)

MnO\(_{4(aq)}^{-1}\) + 8H\(^+ _{(aq)}\) + Y → Mn\(^{2+} _{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
Y in the equation above represents

2e^-

3e^-

5e^-

7e^-

We want the number of charges on both sides of the equation to be the same.

i.e. number of charges on the reactant side = the number of charges on the product side.

-1 + 8 = 2

7 = 2

difference in number of charges = 5, hence Y = 5.