The two most common oxidation states are Cr(III) and Cr(VI). Balance the following oxidation-reduction equations that involve one of these oxidation states of chromium. Assume that the reactions occur in acidic solution.
A. Cr(s) + O2 (g) + H^+ <-> Cr^3+
B. Fe^2+ (aq) + Cr2O7^2- (aq) <-> Fe^3+ (aq) + Cr^3+ (aq)
C. BrO3- (aq) +Cr^3+ (aq) <-> Br2 (aq) + HCrO4- (aq)
I know the steps of solving one of these but I can't seem to get these equations to fit. Please help
All I can do is to balance these for you and that won't help you at all. If you want to show what you have, especially for the one you know, perhaps we can help. Here is a site that may help, also.
http://www.chemteam.info/Redox/Redox.html
In 1, Cr goes from zero on the left to 3+ on the right. Oxygen is zero on the left
I just know the 6 steps to balance:
1. identification
2. balance mass
3. balance oxygen
4. balance hydrogen
5. balance charge
6. combine
To balance oxidation-reduction (redox) equations, you need to follow a systematic approach. Here's how you can balance each equation:
A. Cr(s) + O2 (g) + H+ → Cr^3+
First, identify the oxidation and reduction half-reactions:
Oxidation: Cr(s) → Cr^3+
Reduction: O2 (g) + H+ → Cr^3+
Now, balance the atoms other than H and O in both half-reactions:
Oxidation: Cr(s) → Cr^3+
Reduction: O2 (g) + 4H+ → 2Cr^3+
Next, balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen:
Oxidation: Cr(s) → Cr^3+
Reduction: O2 (g) + 4H+ → 2Cr^3+ + 2H2O
Now balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen:
Oxidation: Cr(s) + 3H2O → Cr^3+ + 6H+
Reduction: O2 (g) + 4H+ → 2Cr^3+ + 2H2O
Finally, balance the charge by adding electrons (e-) to the side that needs more charge:
Oxidation: Cr(s) + 3H2O → Cr^3+ + 6H+ + 3e-
Reduction: O2 (g) + 4H+ + 4e- → 2Cr^3+ + 2H2O
Now multiply the half-reactions by appropriate coefficients to equalize the transferred electrons on both sides:
Oxidation: 2Cr(s) + 6H2O → 2Cr^3+ + 12H+ + 6e-
Reduction: 3O2 (g) + 12H+ + 12e- → 6Cr^3+ + 6H2O
Finally, add the two half-reactions together and cancel out any common species:
2Cr(s) + 6H2O + 3O2 (g) + 12H+ → 2Cr^3+ + 12H+ + 6e- + 6Cr^3+ + 6H2O
Cancel out the common species and simplify:
2Cr(s) + 3O2 (g) + 6e- → 6Cr^3+
B. Fe^2+ (aq) + Cr2O7^2- (aq) → Fe^3+ (aq) + Cr^3+ (aq)
For this equation, follow the same steps as above to balance the equation. First, split the reaction into two half-reactions:
Oxidation: Cr2O7^2- (aq) → Cr^3+ (aq)
Reduction: Fe^2+ (aq) → Fe^3+ (aq)
Balance the atoms except for H and O:
Oxidation: Cr2O7^2- (aq) → 2Cr^3+ (aq)
Reduction: Fe^2+ (aq) → Fe^3+ (aq)
Balance the oxygen atoms:
Oxidation: Cr2O7^2- (aq) → 2Cr^3+ (aq) + 7H2O
Reduction: Fe^2+ (aq) → Fe^3+ (aq)
Balance the hydrogen atoms:
Oxidation: Cr2O7^2- (aq) → 2Cr^3+ (aq) + 7H2O
Reduction: Fe^2+ (aq) + 2H+ → Fe^3+ (aq)
Balance the charge:
Oxidation: Cr2O7^2- (aq) + 14H+ → 2Cr^3+ (aq) + 7H2O
Reduction: Fe^2+ (aq) + 2H+ → Fe^3+ (aq)
Multiply each half-reaction by appropriate coefficients to equalize the transferred electrons:
Oxidation: 6Cr2O7^2- (aq) + 42H+ → 12Cr^3+ (aq) + 21H2O + 12e-
Reduction: 24Fe^2+ (aq) + 48H+ + 12e- → 24Fe^3+ (aq)
Add the two half-reactions together and cancel out common species:
6Cr2O7^2- (aq) + 24Fe^2+ (aq) + 42H+ → 12Cr^3+ (aq) + 21H2O + 24Fe^3+
Simplify the equation if needed.
C. BrO3- (aq) +Cr^3+ (aq) → Br2 (aq) + HCrO4- (aq)
Follow the same steps as above to balance this equation. First, split the reaction into half-reactions:
Oxidation: Cr^3+ (aq) → CrO4^2- (aq)
Reduction: BrO3- (aq) → Br2 (aq)
Balance the atoms:
Oxidation: Cr^3+ (aq) → CrO4^2- (aq)
Reduction: BrO3- (aq) → 3Br2 (aq)
Balance the oxygen atoms:
Oxidation: Cr^3+ (aq) → CrO4^2- (aq)
Reduction: BrO3- (aq) → 3Br2 (aq) + 6 H2O
Balance the charge:
Oxidation: 6Cr^3+ (aq) → 6CrO4^2- (aq)
Reduction: BrO3- (aq) + 6H+ → 3Br2 (aq) + 6H2O
Multiply each half-reaction by appropriate coefficients to equalize the transferred electrons:
Oxidation: 6Cr^3+ (aq) → 6CrO4^2- (aq) + 6e-
Reduction: 2BrO3- (aq) + 12H+ + 6e- → 6Br2 (aq) + 6H2O
Add the two half-reactions together and cancel out common species:
6Cr^3+ (aq) + 2BrO3- (aq) +12H+ → 6CrO4^2- (aq) + 6Br2 (aq) + 6H2O
Simplify the equation if necessary.