Please Help!!
I don't know how to balance this redox equation! I've been trying to solve this for hours and I'm getting nowhere
Fe(s) + HClO4 (aq) --> H (g) + FeClO4
I tried balancing and I got these
Fe(s) + HClO4 (aq) --> H2 (g) + Fe(ClO4)3
and
2Fe(s) + 6HClO4 (aq) --> 3H2 (g) + 2Fe(ClO4)3
and
Fe(s) + HClO4 (aq) --> H (g) + FeClO4
but it was incorrect. I don't know how to solve this problem. I'd really appreciate the help PLEASE!
Maybe my products are wrong?? I don't know...please I'd appreciate any help
I understand your frustration. Balancing redox equations can be challenging, but with a systematic approach, you can find the correct answer.
To balance redox equations, you need to follow these steps:
1. Assign oxidation numbers: Start by assigning oxidation numbers to the atoms in the equation. Since Fe is a solid, its oxidation number is 0. The oxidation number of H is +1, and the oxidation number of Cl in HClO4 is +7.
2. Identify the oxidized and reduced species: Determine which species is undergoing oxidation and which is being reduced. In this case, Fe is being oxidized from an oxidation number of 0 to +2, and H in HClO4 is being reduced from +7 to 0.
3. Write two half-equations: Split the overall equation into two half-equations, one for the oxidation and one for the reduction. In this case, the oxidation half-equation is:
Fe(s) -> Fe^2+ (aq) + 2e^-
And the reduction half-equation is:
H^+ (aq) + 4e^- -> H2 (g)
4. Balance the atoms: Balance the atoms in each half-equation, excluding O and H. In the oxidation half-equation, there is only one Fe on both sides, so it is already balanced. In the reduction half-equation, you need to balance the H atoms by adding 4 H^+ ions on the left-hand side.
5. Balance the charges: Adjust the number of electrons (e^-) so that the total charge is balanced on both sides. In the oxidation half-equation, there are 2 electrons on the right-hand side, so it is balanced. In the reduction half-equation, there are 4 electrons on the left-hand side, so no further adjustment is needed.
6. Combine the half-equations: Multiply each half-equation by the small whole number, so the number of electrons canceled. In this case, you need to multiply the oxidation half-equation by 4 and the reduction half-equation by 2 to make the number of electrons equal.
The balanced half-equations are:
4Fe(s) -> 4Fe^2+ (aq) + 8e^-
8H^+ (aq) + 8e^- -> 4H2 (g)
7. Cancel out common species: Since both half-equations have 8e^- and 8H^+, you can cancel them out, simplifying the equation.
The simplified balanced equation is:
4Fe(s) + 8HClO4(aq) -> 4Fe^2+(aq) + 4H2O(l) + 8ClO4^-(aq)
Therefore, the correct balanced redox equation is:
4Fe(s) + 8HClO4(aq) -> 4Fe^2+(aq) + 4H2O(l) + 8ClO4^-(aq)
I hope this explanation helps you understand the process of balancing redox equations. If you have any further questions, please feel free to ask!
H should be H2.
FeClO4 should be Fe(ClO4)2
Fe + 2HClO4 ==> Fe(ClO4)2 + H2
This equation you wrote is balanced but the predominate product isn't Fe(ClO4)3.
2Fe(s) + 6HClO4 (aq) --> 3H2 (g) + 2Fe(ClO4)3