MnO4 ^- + C2O2 ^2- = MnO2 + CO3 ^2-
how would you balance this in basic solution? I keep getting the wrong answer.
I think theres a mistake in the book its supposed to be C2O4^2-.
and it says it says the answer's supposed to be
2MnO4^- + 3C2O4^2- + 4OH- => 2MnO2 + 6CO3^2- + 2H2O but I didn't get that.
To balance the given equation in basic solution, follow these step-by-step instructions:
Step 1: Write down the unbalanced equation:
MnO4^-(aq) + C2O2^2-(aq) = MnO2(s) + CO3^2-(aq)
Step 2: Identify the elements that are unbalanced. In this case, we need to balance manganese (Mn), carbon (C), and oxygen (O).
Step 3: Start by balancing the atoms that appear in only one compound on each side of the equation. In this case, we can balance manganese (Mn) first.
MnO4^-(aq) + C2O2^2-(aq) = MnO2(s) + CO3^2-(aq)
Step 4: Balance the oxygen (O) atoms by adding water (H2O) molecules to the appropriate side of the equation. The number of water molecules should be equal to the difference in the number of oxygen atoms.
MnO4^-(aq) + C2O2^2-(aq) = MnO2(s) + CO3^2-(aq) + H2O
The equation is now:
MnO4^-(aq) + C2O2^2-(aq) = MnO2(s) + CO3^2-(aq) + H2O
Step 5: Balance the hydrogen (H) atoms by adding hydrogen ions (H+) to the appropriate side of the equation. The number of hydrogen ions should be equal to the difference in the number of hydrogen atoms.
MnO4^-(aq) + C2O2^2-(aq) + 2H2O = MnO2(s) + CO3^2-(aq) + H2O + 2H+
The equation is now:
MnO4^-(aq) + C2O2^2-(aq) + 2H2O = MnO2(s) + CO3^2-(aq) + H2O + 2H+
Step 6: Balance the charges by adding electrons (e^-) as necessary. Since this is a basic solution, the electrons are added to the side where the reduction half-reaction takes place. In this case, the MnO4^- ion is reduced to MnO2. The MnO4^- ion has a charge of -1, and MnO2 has a charge of 0, so we need to balance the charges by adding 5 electrons.
MnO4^-(aq) + 5e^- + C2O2^2-(aq) + 2H2O = MnO2(s) + CO3^2-(aq) + H2O + 2H+
The equation is now:
MnO4^-(aq) + 5e^- + C2O2^2-(aq) + 2H2O = MnO2(s) + CO3^2-(aq) + H2O + 2H+
Step 7: Balance the carbon (C) atoms by adding carbonate (CO3^2-) ions to the appropriate side of the equation. The number of carbonate ions should be equal to the difference in the number of carbon atoms.
MnO4^-(aq) + 5e^- + C2O2^2-(aq) + 2H2O = MnO2(s) + CO3^2-(aq) + H2O + 2H+
The equation is now:
MnO4^-(aq) + 5e^- + C2O2^2-(aq) + 2H2O = MnO2(s) + CO3^2-(aq) + H2O + 2H+
Now, the equation is balanced in basic solution.
To balance this chemical equation in a basic solution, follow these steps:
Step 1: Write the unbalanced equation:
MnO4^- + C2O2^2- = MnO2 + CO3^2-
Step 2: Separate the equation into half-reactions, one for the oxidation half (losing electrons) and one for the reduction half (gaining electrons).
Oxidation half-reaction: MnO4^- → MnO2
Reduction half-reaction: C2O2^2- → CO3^2-
Step 3: Balance the atoms except for hydrogen and oxygen in each half-reaction.
Oxidation half-reaction: MnO4^- → MnO2 + 4e^-
Reduction half-reaction: C2O2^2- + 3e^- → CO3^2-
Step 4: Balance the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 4.
Oxidation half-reaction: 3MnO4^- → 3MnO2 + 12e^-
Reduction half-reaction: 4C2O2^2- + 12e^- → 4CO3^2-
Step 5: Multiply each half-reaction by a factor to equalize the number of electrons transferred. In this case, the least common multiple of 12 and 12 is 12, so multiply the oxidation half-reaction by 1 and the reduction half-reaction by 1.
Oxidation half-reaction: 3MnO4^- → 3MnO2 + 12e^-
Reduction half-reaction: 4C2O2^2- + 12e^- → 4CO3^2-
Step 6: Combine the half-reactions, canceling out the electrons to obtain the balanced overall equation.
3MnO4^- + 4C2O2^2- → 3MnO2 + 4CO3^2-
Since it is a basic solution, you need to balance the charges by adding OH^- ions to both sides of the equation.
3MnO4^- + 4C2O2^2- + 6OH^- → 3MnO2 + 4CO3^2- + 3H2O
The balanced equation in basic solution is:
3MnO4^- + 4C2O2^2- + 6OH^- → 3MnO2 + 4CO3^2- + 3H2O
Are you sure this is the right equation? Here is the so-called C2O2^2-
8OH^- + C2O2^2- ==> 2CO3^2- + 4H2O + 6e