Can you check these? And I need help on one.
Oxidation Number Method of Balancing Redox Reactions
1)Neutral Solution
Sn^2+ + Fe^3+ ----> Fe^2+ + Sn^4+
Balanced equation
Sn^2+ + 2Fe^3+ ----> 2Fe^2+ + Sn^4+
2)Acidic Solution
Fe^2+ + Cr2O7^2- ---> Cr^3+ + Fe^3+
Balanced equation
14H^+ + 6Fe^2+ + Cr2O7^2- ---> 2Cr^3+ + 6Fe^3+ + 7H2O
3) Basic Solution
Mno4^- + SO3^2- ---> SO4^2- + MnO2
Balanced Equation
I can't figure out how to do this one.
First -
2MnO4 + 3So3^2- ---> 3SO4^2- + 2MnO2
Then I added H2O on one side to help balance the O's. Then I had to add an H+ to the reactants side to balance out the H's. Since it was a basic solution I had to add OH- to the answer side. But then I had to add it to the other one as well.
I have -
OH- + H- + 2MnO4 + 3So3^2- ---> 3SO4^2- + 2MnO2 + H2O + OH-
OH- + H- becomes H2O...the H's are unbalanced and I don't know how to keep them balanced while making sure it's basic.
4) Basic Solution
CH3OH + MnO4^- ---> MnO4^2- + CO3^2-
Balanced Equation
CH3OH + 6MnO4^- ---> 6MnO4^2- + CO3^2-
I don't know how to balance out the H's.
Also, can you please explain to me the half reaction method?
Can you do this question and explain what you did.
Hg + H3IO6^2- ---> Hg2O + IO3^-
Thanks so much for all the help!
Let's check the balanced equations you provided:
1) Neutral Solution
Sn^2+ + 2Fe^3+ ----> 2Fe^2+ + Sn^4+
The balanced equation you provided is correct.
2) Acidic Solution
14H^+ + 6Fe^2+ + Cr2O7^2- ---> 2Cr^3+ + 6Fe^3+ + 7H2O
The balanced equation you provided is correct.
3) Basic Solution
To balance the equation in a basic solution, you need to add OH^- ions to both sides to neutralize the excess H^+ ions. Let's work on it step by step:
2MnO4^- + 3SO3^2- ---> 3SO4^2- + 2MnO2
To balance the O atoms, add H2O to the left side:
2MnO4^- + 3SO3^2- + H2O ---> 3SO4^2- + 2MnO2
Next, balance the H atoms by adding H^+ ions to the right side:
2MnO4^- + 3SO3^2- + H2O ---> 3SO4^2- + 2MnO2 + 2H^+
Now, to balance the charge, add OH^- ions to both sides:
2MnO4^- + 3SO3^2- + H2O + 6OH^- ---> 3SO4^2- + 2MnO2 + 2H^+ + 6OH^-
Simplifying, you get:
2MnO4^- + 3SO3^2- + H2O + 2OH^- ---> 3SO4^2- + 2MnO2 + 2H2O + 2OH^-
Finally, cancel out the water molecules:
2MnO4^- + 3SO3^2- + OH^- ---> 3SO4^2- + 2MnO2 + H2O
So, the balanced equation in basic solution is:
2MnO4^- + 3SO3^2- + OH^- ---> 3SO4^2- + 2MnO2 + H2O
4) Basic Solution
CH3OH + 6MnO4^- ---> 6MnO4^2- + CO3^2-
The balanced equation you provided is correct.
Now, let's talk about the half-reaction method. In this method, you separate the oxidation and reduction half-reactions and balance them individually. Here's how you can balance the given reaction using the half-reaction method:
Hg + H3IO6^2- ---> Hg2O + IO3^-
First, identify the oxidation and reduction half-reactions. In this case:
Oxidation:
H3IO6^2- ---> IO3^- (loss of electrons)
Reduction:
Hg ---> Hg2O (gain of electrons)
Now, balance each half-reaction separately:
Oxidation:
H3IO6^2- ---> 2IO3^- + 4H+ + 3e-
Reduction:
Hg + 2H2O ---> Hg2O + 4H+ + 4e-
Multiply the half-reactions by appropriate coefficients to make the number of electrons equal. In this case, multiply the oxidation half-reaction by 4 and the reduction half-reaction by 3.
4(H3IO6^2- ---> 2IO3^- + 4H+ + 3e-)
3(Hg + 2H2O ---> Hg2O + 4H+ + 4e-)
Now, add the balanced half-reactions together:
4H3IO6^2- + 3Hg + 6H2O ---> 2IO3^- + Hg2O + 8H+ + 9e-
Simplify and cancel out common terms:
4H3IO6^2- + 3Hg + 6H2O ---> 2IO3^- + Hg2O + 8H+ + 9e-
So, the balanced equation using the half-reaction method is:
4H3IO6^2- + 3Hg + 6H2O ---> 2IO3^- + Hg2O + 8H+ + 9e-
I hope this helps! Let me know if you have any further questions.