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. Fe(s) + 8H+(aq) + MnO4–(aq)Mn2+(aq) + 4H2O(l) + Fe2+(aq)
B. 5Fe(s) + 16H+(aq) + 2MnO4–(aq)2Mn2+(aq) + 8H2O(l) + 5Fe2+(aq)
C. Fe(s)+8H+(aq)+MnO4–(aq)+Pt2+(aq)Mn2+(aq)+4H2O(l)+Fe2+(aq)+Pt(s)
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. Mn2+(aq) + 4H2O(l)5e– + 8H+(aq) + MnO4–(aq)
B. Fe(s)Fe2+(aq) + 2e–
C. Fe2+(aq) + 2e–Fe(s)
D. 5e– + 8H+(aq)+MnO4–(aq)Mn2+(aq) +4H2O(l)
E. Pt(s)Pt2+ + 2e–
B and D
To determine the correct balanced chemical equation and the reaction occurring at the cathode in the given electrochemical cell, we need to understand the conventions for writing the half-reactions in an acidic solution.
The cathode is the electrode where reduction occurs, meaning reduction half-reaction takes place. In the given options, only option B. Fe(s)Fe2+(aq) + 2e– represents a reduction half-reaction.
Now, let's determine the balanced chemical equation in an acidic solution. We'll consider the oxidation half-reaction and the reduction half-reaction:
Oxidation half-reaction:
Fe(s) → Fe2+(aq) + 2e–
Reduction half-reaction:
MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)
To balance these two half-reactions, we need to multiply the oxidation half-reaction by 5 and multiply the reduction half-reaction by 2:
5Fe(s) → 5Fe2+(aq) + 10e–
2MnO4–(aq) + 16H+(aq) + 10e– → 2Mn2+(aq) + 8H2O(l)
Now, we can add these two equations together to form the balanced chemical equation:
5Fe(s) + 2MnO4–(aq) + 16H+(aq) → 5Fe2+(aq) + 2Mn2+(aq) + 8H2O(l)
Therefore, the correct balanced chemical equation in an acidic solution is B. 5Fe(s) + 16H+(aq) + 2MnO4–(aq) → 2Mn2+(aq) + 8H2O(l) + 5Fe2+(aq).
The reaction occurring at the cathode is represented by B. Fe(s)Fe2+(aq) + 2e–, where Fe2+ ions are reduced to Fe atoms.