For the following electrochemical cell:

Fe(s) / Fe2+(aq) // MnO4–(aq) Mn2+(aq) / Pt(s)
Which letter corresponds to the correct balanced chemical equation in an acidic solution?

A. 5Fe(s) + 16H+(aq) + 2MnO4–(aq)--> 2Mn2+(aq) + 8H2O(l) + 5Fe2+(aq)
B. Fe(s)+8H+(aq)+MnO4–(aq)+Pt2+(aq)---> Mn2+(aq)+4H2O(l)+Fe2+(aq)+Pt(s)
C. Fe(s) + 8H+(aq) + MnO4–(aq)--->Mn2+(aq) + 4H2O(l) + Fe2+(aq)
D. 2Mn2+(aq) + 8H2O(l) + 5Fe2+(aq)--> 5Fe(s) + 16H+(aq) + 2MnO4–(aq)
E. Mn2+(aq)+4H2O(l)+Fe2+(aq)+Pt(s)----> Fe(s)+8H+(aq)+MnO4–(aq)+Pt2+(aq)

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

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I read what you said a while ago but i still don't know how to go about the first part. For some reason when i add the reactions I get A

Is the second part B?

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.

To determine the correct balanced chemical equation in an acidic solution for the given electrochemical cell, we need to follow a systematic approach. Here's how you can solve it:

1. Identify the half-reactions: In this cell, we have two half-reactions happening at each electrode. The half-reactions are:
a) Oxidation half-reaction (anode): Fe(s) → Fe2+(aq) + 2e–
b) Reduction half-reaction (cathode): MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

2. Balance the half-reactions: Start by balancing the atoms in each half-reaction, excluding hydrogen and oxygen. In the oxidation half-reaction, Fe is already balanced. In the reduction half-reaction, we have one Mn atom on each side of the equation. To balance hydrogen, add 8H+ ions to the oxidation half-reaction. To balance oxygen, add 4H2O molecules to the oxidation half-reaction.

The balanced half-reactions become:
a) Oxidation half-reaction: Fe(s) → Fe2+(aq) + 2e– + 8H+(aq)
b) Reduction half-reaction: MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

3. Determine the number of electrons transferred: By comparing the number of electrons in both half-reactions, we see that 5 electrons are transferred in the reduction half-reaction, which matches with the oxidation half-reaction.

4. Multiply the half-reactions to balance the number of electrons: Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the number of electrons transferred.

The balanced half-reactions become:
a) Oxidation half-reaction: 5Fe(s) → 5Fe2+(aq) + 10e– + 40H+(aq)
b) Reduction half-reaction: 2MnO4–(aq) + 16H+(aq) + 10e– → 2Mn2+(aq) + 8H2O(l)

5. Add the balanced half-reactions: Add the balanced half-reactions together to obtain the overall balanced chemical equation.

The correct balanced chemical equation in an acidic solution is:
5Fe(s) + 2MnO4–(aq) + 16H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5Fe2+(aq)

Therefore, the correct answer for the first part is option A.

Now, let's move on to the second part:

To determine the reaction occurring at the cathode, we need to look at the reduction half-reaction. The reduction half-reaction is the one that occurs at the cathode. In this case, the reduction half-reaction is:

Reduction half-reaction: MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)

So, the correct answer for the second part is option A.

I hope this clears up any confusion you had. Let me know if you have any further questions!