Of course it's A for part 1. Remember how to separate the equations into their two half cells, balance each independently, then multiply by whatever numbers are necessary to make the loss and gain of electrons equal and add the two. Those equations will balance every time.
For part 2, you only need to remember the one definition. Oxidation occurs at the ANODE. So reduction must occur at the cathode (which is what the question asks). Reduction is the gain of electrons, right. Did answer B gain electrons? Nope. Doesn't look like it gained to me. In fact, I see ONLY ONE equation there that gained electrons. So that must the be the right answer. It's past my bed time. I'll stay a minute or two longer to see if this is clear. Please post as soon as you can to let me know
So part 2 is D and part 1 is A jus making that clear...
Yes, part 1 is A.
But part 2 is not D. C'mon. I said I saw only one equation that gained electrons. B, your first guess, has electrons on the right hand side which means it is losing electrons which means that is oxidation. Answer D has electrons on the right side which means it is losing electrons and is oxidation. There is only one equation there with electrons on the left side which means that reaction is gaining electrons, that means it is reduction, and that means it is occurring at the cathode.
What reaction is occurring at the cathode?
A. 5e– + 8H+(aq)+MnO4–(aq)---> Mn2+(aq) +4H2O(l)
B. Fe(s)---> Fe2+(aq) + 2e–
C. Mn2+(aq) + 4H2O(l)----> 5e– + 8H+(aq) + MnO4–(aq)
D. Fe2+(aq) + 2e– --->Fe(s)
E. Pt(s)---> Pt2+ + 2e
D doesn't have electrons on the right side and neither does A
OK. I think I see part of the problem. I goofed in my earlier response(s). I said there was only one reaction with electrons on the left side, or a gain of electrons.As you have correctly pointed out, answers A and D have electrons on the left side; therefore, both answers A and D are reductions Of the two answers, we want the reaction occurring at the cathode. That is reduction and that is a gain of electrons. If Fe==>Fe^+2 + 2e is occurring at the anode because it is oxidation, then MnO4^- must be the reduction occurring at the cathode. So A must be the correct answer. As an added measure of assurance that this is the correct answer, look at the electrochemical notation of the cell in the original problem. You see Fe is on the left and MnO4^- is on the right. Following convention, the anode is on the left and the cathode on the right in spontaneous cells. I hope this helps clear things up.
I see I just took on a new screen name.