Balance the following redox reaction occurring in acidic solution.
H+ + Cr --> H2 + Cr3+
H+ --> H2 do I add H+ to this side since H2 = 0? Also what about electrons.
Cr -> Cr3+ + 3e-
Hope I make sense
H^+ ==> H2
First balance the H atoms so we are looking at the same number on each side. That's about the only way to balance redox equations.You do that so you can talk about total charges on each side.
2H^+ ==> H2
H has changed from +2(+1 each) to zero; therefore add 2e to the left.
2H^+ + 2e ==> H2
Cr is correct at Cr==> Cr^3+ + 3e.
Then you multiply the H equation by 3 and the Cr equation by 2 and add them. Cancel the electrons (there will be 6 on each side) and you have it.
Yes, you are on the right track. Let's balance the redox reaction step-by-step:
1. Write the unbalanced equation:
H+ + Cr --> H2 + Cr3+
2. Balance the atoms other than hydrogen and oxygen:
Since the number of chromium (Cr) atoms is already balanced on both sides, we move on to balance hydrogen atoms.
H+ + Cr --> H2 + Cr3+
We have 2 hydrogen atoms on the right side and only 1 on the left side. To balance the hydrogen atoms, we need to add H+ to the left side. The equation becomes:
2H+ + Cr --> H2 + Cr3+
3. Balance the charges:
Now let's balance the charges by adding electrons (e-) to the appropriate side of the equation. In this case, the chromium (Cr) goes from a charge of 0 to +3, so it gains 3 electrons.
H+ + Cr --> H2 + Cr3+ + 3e-
4. Verify the balance of atoms and charges on both sides:
On the left side, we have 2 hydrogen atoms, 1 chromium atom, and a total charge of +2 (2H+). On the right side, we have 2 hydrogen atoms, 1 chromium atom with a +3 charge (Cr3+), and a total charge of +2 (2H+ + 3e-).
Both sides have the same number of atoms and charges, so the equation is now balanced:
2H+ + Cr --> H2 + Cr3+ + 3e-
Yes, you are correct about the half-reaction for the reduction of Cr. The oxidation number of Cr changes from 0 to +3, so it loses 3 electrons in the process:
Cr → Cr3+ + 3e-
For the half-reaction representing the reduction of H+, you have correctly identified that H+ is reduced to H2. Since H2 has an oxidation state of 0, you do not need to add any H+ to the equation.
Now, to balance the overall redox reaction, you need to ensure that the number of electrons lost in the oxidation half-reaction (Cr) is equal to the number of electrons gained in the reduction half-reaction (H+).
In this case, you can multiply the reduction half-reaction by 1 (to balance the electrons) and the oxidation half-reaction by 3 (to match the number of electrons gained in the reduction):
3Cr → 3Cr3+ + 9e-
2H+ + 3e- → H2
Now, if you add these two reactions together, the electrons on both sides will cancel out, and the equation will be balanced:
3Cr + 2H+ → 3Cr3+ + H2
Therefore, to balance the redox reaction occurring in acidic solution:
3Cr + 2H+ → 3Cr3+ + H2