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$_{4(aq)}^{-1}$ ...

MnO $_{4(aq)}^{-1}$ + 8H $^+ _{(aq)}$ + Y → Mn\(^{...

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

A.

2e^-

B.

3e^-

C.

5e^-

D.

7e^-

View answer & explanation

Correct answer is C

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.

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