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
In an acid solution, MnO4^- is reduced to Mn^+2.
MnO4^- + 5e + 8H^+ ==>Mn^+2 + 4H2O
Well, let's break it down! In the reaction, we have Fe2+ being oxidized to Fe3+, while MnO4- is being reduced to Mn2+.
Now, we need to find the reduction half-reaction, which shows the species being reduced. In this case, it's MnO4-. So, the reduction half-reaction would be:
b. MnO4–(aq) + 8H+(aq) + 5e– --> Mn2+(aq) + 4H2O
Remember, reduction is the "gain" of electrons, and in this case, MnO4- is gaining 5 electrons to become Mn2+. It's like MnO4- is saying "Hey, can I borrow 5 electrons? I want to feel complete!"
So, the answer is b. MnO4–(aq) + 8H+(aq) + 5e– --> Mn2+(aq) + 4H2O.
The reduction half-reaction for the reaction between iron(II) sulfate and potassium permanganate in a sulfuric acid solution is option 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 that undergoes reduction and calculate the change in oxidation state for that species.
In the given balanced equation:
5Fe2+(aq) + MnO4–(aq) + 8H+(aq) --> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
We can see that Fe2+ is being oxidized to Fe3+, which means that reduction is occurring for the manganese species, MnO4–(aq), and hydrogen ions, H+(aq).
Let's focus on the manganese species, MnO4–(aq). The starting state of manganese is +7 in MnO4– and it is reduced to +2 in Mn2+. We can see that the oxidation state of manganese decreases by 5.
Therefore, the reduction half-reaction for the reaction is:
MnO4–(aq) + 5e– --> Mn2+(aq)
Answer: c. MnO4–(aq) + 5e– --> Mn2+(aq) + SO42–(aq)