What is the reduction half-reaction for the reaction between iron(II) sulfate and potassium permanganate in a sulfuric acid solution?

5Fe2+(aq) + MnO4–(aq) + 8H+(aq) --> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

a. MnO4– + 3e– --> Mn2+(aq) + 4H2O
b. MnO4–(aq) + 8H+(aq) + 5e– ---> Mn2+(aq) + 4H2O
c. MnO4–(aq) + 5e– ---> Mn2+(aq) + SO42–(aq)
d. Fe2+(aq) + 2e– --->Fe(s)
e. 5Fe2+(aq) ---> 5Fe3+(aq) + 5e–

The reduction half-reaction for the reaction between iron(II) sulfate and potassium permanganate in a sulfuric acid solution is:

e. 5Fe2+(aq) ---> 5Fe3+(aq) + 5e–

To determine the reduction half-reaction for the reaction between iron(II) sulfate and potassium permanganate in a sulfuric acid solution, we need to identify the species being reduced.

In the given balanced equation:

5Fe2+(aq) + MnO4–(aq) + 8H+(aq) --> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

The Fe2+ ions are being oxidized to Fe3+ ions, while the MnO4- ions are being reduced to Mn2+ ions.

The reduction half-reaction involves the gain of electrons (reduction), so we need to determine which choice represents the gain of electrons for the MnO4- ions.

Looking at the options:
a. MnO4– + 3e– --> Mn2+(aq) + 4H2O
b. MnO4–(aq) + 8H+(aq) + 5e– ---> Mn2+(aq) + 4H2O
c. MnO4–(aq) + 5e– ---> Mn2+(aq) + SO42–(aq)
d. Fe2+(aq) + 2e– --->Fe(s)
e. 5Fe2+(aq) -----> 5Fe3+(aq) + 5e–

Option b represents the reduction half-reaction for the MnO4–(aq) ions. It shows the gain of 5 electrons, which matches the balanced equation. Therefore, the correct answer is b. MnO4–(aq) + 8H+(aq) + 5e– ---> Mn2+(aq) + 4H2O.

did you ever get an answer?

Answered above.