Consider the following reaction and select the FALSE statement.

Question

Consider the following reaction and select the FALSE statement.

10 I^-(aq) + 16 H⁺(aq) + 2 MnO₄^-(aq) --> 5 I₂(s) + 2 Mn²⁺(aq) + 8 H₂O(l)

1) The iodine ion is the reducing agent

2) This is a redox reaction.

3) The permanganate ion is the oxidizing agent

4) In the reaction as written, ten moles of electrons are transferred from the oxidizing agent to the reducing agent.

5) The hydrogen ion is neither reduced nor oxidized.

I am between choices 4 and 5, but I may be wrong and it might be another choice. I thought they were all true, but I know that can't be the case since there is one right answer. Can someone please provide an explanation to the correct answer and why the other choices (especially 4/ 5) are wrong?

Answer 1

I^- goes from -1 to zero. It has been oxidized and is the reducing agent. So (1) is correct

This is indeed a redox so (2) is correct

The Mn in MnO₄^- goes +7 to +2. It has been reduced and is the oxidizing agent. So (3) is sort of correct but its actually Mn as opposed to permanganate

Since there are ten I^- on left and 5 I₂ on right, that’s a transfer of 10 electrons. Same for Mn which has 2 at +7 going to 2 at +2 for a total of 10 electrons. So (4) is correct

The hydrogen on the left is +1 and on the right (in H2O) is is also +1 so (5) is correct.

Although a little sneaky, #3 is the least correct answer as explained above.