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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First Problem: ; would be C?
Second problem: would that be A because positively charged ions are coming from it
Did you type the question correctly? You have one question and two set of answers.
yea i just forgot the space in between sorry
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
I see that now. I just didn't look closely enough.
Write the two half reactions. I don't think you can come up with C. In fact, I think C is and OLD OLD trick. All the atoms balance, don't they? Yes, Very good. BUT, do the charges balance? No, I don't think so. Is the electron transfer lost equal to those gained? No, I don't think so. The charge is +7 on the left and +4 on the right. No good. The electron transfer is 2 for Fe going to Fe^+2 and 5 for MnO4^- at +7 going to Mn^+2 at +2. So the electrons don't balance either. Try this.
Fe==> Fe^+2 + 2e
MnO4^- + 8H^+ 5e ==> Mn^+2 + 4H2O
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Now make the electrons lost equal those gained and add the two equations. You will get the correct balanced chemical equation. (This takes care of electrons transfer and charge balance at the same time.)
For the first problem, to determine the correct balanced chemical equation in an acidic solution, we need to consider the balancing of reactants and products, as well as the conservation of mass and charge. Let's analyze each option:
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)
To balance the redox equation, let's assign oxidation states:
Fe(s) --> Fe2+(aq)
MnO4–(aq) --> Mn2+(aq)
Option A is the correct answer because it correctly balances the oxidation and reduction half-reactions.
Now, for the second problem, we need to identify the reaction occurring at the cathode. The cathode is where reduction occurs. Let's examine the options:
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
Option B is the correct answer because it describes the reduction half-reaction at the cathode: Fe(s) --> Fe2+(aq) + 2e–. The cathode is where reduction occurs, so Fe(s) is being reduced to Fe2+(aq).