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!