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
Y in the equation above represents
2e-
3e-
5e-
7e-
Explicació
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.