Question

MnO4- (aq) + SbH3 (aq) → MnO2 (s) + Sb (s)

What is being oxidized?

Answer 1

Well, separate the compounds:
If Mn + O4 = -1, and the charge on each O is -2, then Mn + (-2)4 = -1 and Mn = -1 + 8 or 7.

Thus, Mn starts out with an initial charge of 7.

If Sb + 3H = 0, then Sb + 3 = 0 and Sb has a charge of -3.

Thus, Sb starts out with a charge of -3.

In the products, Sb is an element and therefore ends up with a charge of 0.

In the products, MnO2 = 0, and 2(-2) + Mn = 0. Therefore, Mn has a final charge of 4.

So, Mn goes from a charge of 7 to 4. Its oxidation number is going down, and it is therefore GAINING electrons. Because reduction is a gain of electrons, Mn is being reduced.

Likewise, Sb starts out with a charge of -3 and ends with a charge of 0. Its oxidation number is going up, and so therefore Sb is LOSING electrons. Because oxidation is the loss of electrons, Sb is being oxidized.

Therefore, Sb is being oxidized. I hope this helps :)

Answer 2

Thanks Your Answer saved me!