Fe2+ is oxidized to Fe3+ and CrO4is reduced to Cr3+ using an unknown sample containing Fe2+ and 0.04322M K2CrO4 solution in a redox titration. I first need to balance the formula, but I don't know what to do with the potassium.
The simple thing to do is to ignore it. It's a spectator ion. The rxn is between the chromate ion and the ferrous (FeII) ion.
The actual question is: A sample is analyzed to determine its iron content (as Fe2+0 via a redox titration with potassium chromate as the titrant. In the titration Fe2+ is oxidized to Fe3+ and CrO4(2-) is reduced to Cr3+. What is the percent by mass of iron in the sample if 0.9087 g of the sample required 45.68 mL of a 0.04322 M K2CrO4 solution to reach the endpoint?
I figured out moles of the titrant. I know that I need to balance the reaction by using the half-reaction method. I just don't know what the the second half looks like. I have:
Fe2+ goes to Fe3+ and e-
(some number of e-) and 8H+ and KCrO4 goes to Cr3+ and 4H2O and (2K??)
So I'm a little lost as to what to do next...
Thanks!
When balancing the redox reaction equation, you should only focus on balancing the atoms involved in the oxidation-reduction process.
To balance the equation between Fe2+ (iron(II)) and Fe3+ (iron(III)), follow these steps:
1. Write down the unbalanced equation:
Fe2+(aq) + CrO4^2-(aq) → Fe3+(aq) + Cr3+(aq)
2. Split the equation into two half-reactions:
Half-reaction for Fe2+:
Fe2+(aq) → Fe3+(aq)
Half-reaction for CrO4^2-:
CrO4^2-(aq) → Cr3+(aq)
3. Balance the atoms other than hydrogen and oxygen in each half-reaction:
For the Fe2+ half-reaction, there is already one Fe atom on each side.
For the CrO4^2- half-reaction, count the number of Cr atoms and oxygen atoms:
Cr: 1 on the left and 1 on the right
O: 4 on the left and 4 on the right
So, the CrO4^2- half-reaction is already balanced for the atoms other than hydrogen and oxygen.
4. Balance the oxygen atoms by adding water molecules (H2O) to the side that needs oxygen.
In the Fe2+ half-reaction, there are 0 oxygen atoms. So, add 3 H2O molecules to the right side:
Fe2+(aq) → Fe3+(aq) + 3H2O(l)
In the CrO4^2- half-reaction, there are 4 oxygen atoms. So, add 4 H2O molecules to the left side:
CrO4^2-(aq) + 4H2O(l) → Cr3+(aq)
5. Balance the hydrogen atoms by adding H+ ions to the side that needs hydrogen.
In the Fe2+ half-reaction, there are 0 hydrogen atoms. So, add 6 H+ ions to the right side:
Fe2+(aq) + 6H+(aq) → Fe3+(aq) + 3H2O(l)
In the CrO4^2- half-reaction, there are 8 hydrogen atoms. So, add 8 H+ ions to the right side:
CrO4^2-(aq) + 4H2O(l) → Cr3+(aq) + 8H+(aq)
6. Balance the charges by adding electrons to the side that needs charge balance.
In the Fe2+ half-reaction, the charge on the left side is +2 and on the right side is +3. So, add 1 electron (e-) to the left side:
Fe2+(aq) + e- → Fe3+(aq) + 3H2O(l)
In the CrO4^2- half-reaction, the charge on the left side is -2 and on the right side is +3. So, add 6 electrons (e-) to the left side:
6e- + CrO4^2-(aq) + 4H2O(l) → Cr3+(aq) + 8H+(aq)
Now, the equation is balanced for both charge and atoms involved in the redox reaction:
Fe2+(aq) + 6H+(aq) + e- → Fe3+(aq) + 3H2O(l)
6e- + CrO4^2-(aq) + 4H2O(l) → Cr3+(aq) + 8H+(aq)
Note that the potassium ions (K+) from K2CrO4 are spectator ions and not involved in the oxidation-reduction process, so you don't need to include them in the balanced equation.
To balance the formula for the redox reaction, you need to assign oxidation numbers to each element and then adjust the coefficients so that the number of atoms and the total charge are equal on both sides of the equation. Let's go step by step:
1. Assign oxidation numbers:
- Oxygen (O) usually has an oxidation number of -2, so in CrO4, each oxygen contributes -8 to the overall charge.
- Since the overall charge of the chromate ion (CrO4) is -2, the oxidation number of chromium (Cr) must be +6.
2. Now, let's focus on the iron (Fe) part:
- It is given that Fe2+ is oxidized to Fe3+. This means that the oxidation number of iron (Fe) increases from +2 to +3.
3. Now, let's balance the equation:
The balanced equation will be:
2Fe2+ + CrO4^2- + 14H+ → 2Fe3+ + Cr^3+ + 7H2O
Note that the potassium ions (K+) from the K2CrO4 do not appear in the balanced equation since they are spectator ions and do not participate in the redox reaction. They are there only to balance the overall charge.
I hope this explanation helps you balance the formula for the redox titration.