Question

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

Answer 1

What do you think and why?
#2. Reduction occurs at the cathode (the + charged electrode).

Answer 2

#1, Fe(s) is on the left, so it's anode; Pt(s) is on the right, it's a cathode...so the rxn should be D.

#2: B because Fe is gaining electrons, making it a reduction=cathode.

Answer 3

Right on the anode and cathode. If D is the correct answer, why doesn't it have Fe in it. There's no Fe in D. And you need an Fe going to Fe^+2 + 2e.

Answer 4

no, there's Fe in D for the first question..Fe(s)+8H+(aq)+MnO4–(aq)+Pt2+(aq)==> Mn2+(aq)+4H2O(l)+Fe2+(aq)+Pt(s)

Answer 5

You're right and I'm wrong. I was looking at the second set of answers and not the first. So Fe is on the left and it is the anode. So now what reaction is occurring at the Pt electrode? Obviously it's the other half cell, which is______?

Answer 6

Pt is functioning as a cathode
- D shows that Pt is cathode and Fe as anode. I don't understand why H+ and water are in the reaction

so my answers
1)D
2)B

Answer 7

D is correct for #1. For #2, isn't the correct answer A? Pt is the cathode (+ electrode where reduction occurs) and that is the MnO4^- reaction going to Mn^+2.

Answer 8

the answer is actually B and A

Answer 9

isn't Pt just functioning as an electrode?

Answer 10

answer is B and A