Balancing Redox Reaction in Basic Solution
CN- + MnO-4 --> CNO- + MnO2
H2O + CN- --> CNO- + 2H +2e
3e + 4H + MnO4- --> MnO2 + 2H2O
My confusion is with adding the OH- please explain
I don't see any OH^- added.
Perhaps you mean CN^- ==> CNO^- in basic solution?
1. CN^- ==> CNO^-
C is +2 on left, +4 on right so we must add 2e to the right.
CN^- ==> CNO^- + 2e
2. Count up the charge on each side. I see -1 on the left and -3 on the right. Add OH^- to the appropriate side.
2OH^- + CN^- ==> CNO^- + 2e
3. Add H2O to balance.
2OH^- + CN^- ==> CNO^- + 2e + H2O
Have you caught on to how I do it? If not I can write up a quickie for you, you can print it out and save it until you have this down pat.
Balance equations by ion-electron method
Step 1. Determine oxidation states of each element and focus on those that change from left side to right side.
Step 2. Multiply, as appropriate so that the number of atoms you are considering are the same; e.g., Cr2O7^2- ==> Cr^3+.
Cr changes from +6 for EACH Cr on the left to +3 for EACH Cr on the right BUT we will NEVER balance the equation by talking about 2 atoms on the left and 1 on the right. Therefore, multiply Cr^3+ on the right by 2 so it looks this way.
Cr2O7^2- ==> 2Cr^3+.
Now we talk about total oxidation numbers. Cr on the left is +12 total and on the right is +6 total which means we add 6e on the left to balance the change in oxidation number.
Step 3. Count the charge on the left and right. Then
(a) if acidic solution, add H^+ to the appropriate side to balance the charge or
(b) if basic solution, add OH^- to the appropriate side to balance the charge.
Step 4. Then add H2O to the appropriate side to balance H and O.
Step 5. Do the other half equation the same way.
Step 6. Multiply each equation by a number to make electrons lost = electrons gained.
Step 7. Add the two multiplied equations.
Step 8. Cancel any ions common to both sides.
Step 9. If you wish to convert to a molecular equation, add cations to negative ions and anions to positive ions on each side of the equation.
Step 10. Check one last time to make sure th equation is balanced three ways.
a. atoms balance.
b. charges balance.
c. electrons lost = electrons gained.
Ah, the balancing act begins! Balancing redox reactions in basic solution can be a bit tricky, but fear not, I'm here to make it more entertaining for you.
To balance the equation and add OH-, you'll follow these steps:
Step 1: Balance the non-hydrogen and non-oxygen elements first. In this case, we start by balancing the charges on both sides of the equation by adding electrons (e-).
CN- + MnO4- → CNO- + MnO2
+ 3e-
Step 2: Balance the oxygen atoms by adding water (H2O) molecules.
CN- + MnO4- → CNO- + MnO2
+ 3e- + 6H2O
Step 3: Balance the hydrogen atoms by adding hydrogen ions (H+).
CN- + MnO4- → CNO- + MnO2
+ 3e- + 6H2O + 6H+
Step 4: For every H+ added, add an equal number of OH- ions to both sides of the equation.
CN- + MnO4- + 4OH- → CNO- + MnO2 + 2H2O + 2H2O
And voila! Your equation is now balanced in basic solution. Easy peasy, lemon squeezy... or should I say OH-some? Keep practicing, and you'll be balancing redox reactions like a pro!
To balance a redox reaction in a basic solution, you need to add hydroxide ions (OH-) to neutralize the excess hydrogen ions (H+) produced during the oxidation-reduction process. Here's how you can add OH- ions to balance the given reaction:
1. Write down the unbalanced equation:
CN- + MnO4- -> CNO- + MnO2
2. Separate the equation into half-reactions: one for the reduction half and the other for the oxidation half.
Reduction half-reaction:
MnO4- -> MnO2
Oxidation half-reaction:
CN- -> CNO-
3. Balance the atoms other than oxygen and hydrogen:
In the reduction half-reaction, you have one manganese (Mn) atom on both sides, so there is no need for further balancing.
In the oxidation half-reaction, you have one carbon (C) atom on both sides, so it is balanced.
4. Balance the oxygen atoms:
In the reduction half-reaction, there are four oxygen (O) atoms on the left side (in MnO4-) and two oxygen atoms on the right side (in MnO2). To balance the oxygen atoms, you need to add water (H2O) molecules to the right side of the equation. So, the reduction half-reaction becomes:
MnO4- -> MnO2 + 2H2O
5. Balance the hydrogen atoms:
In the reduction half-reaction, there are no hydrogen (H) atoms, so it is already balanced.
In the oxidation half-reaction, there are two hydrogen (H) atoms on the right side (in 2H+). To balance the hydrogen atoms, add the same number of OH- ions to the left side of the equation. This will convert the hydrogen ions into water molecules. So, the oxidation half-reaction becomes:
CN- + 2OH- -> CNO- + H2O + 2e
6. Balance the charge on both sides of the half-reactions:
In the reduction half-reaction, the charge is not balanced as there are two electrons (2e-) on the right side and no charge on the left side. To balance the charge, add two more electrons to the left side of the equation. So, the balanced reduction half-reaction becomes:
MnO4- + 2e- -> MnO2 + 2H2O
In the oxidation half-reaction, the charge is already balanced with two electrons (2e-) on the right side and two hydroxide ions (2OH-) on the left side.
7. Multiply the half-reactions by appropriate values to make the number of electrons equal in both half-reactions:
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to make the number of electrons equal on both sides (6e-).
6(CN-) + 6(OH-) -> 6(CNO-) + 3H2O + 6e-
2(MnO4-) + 6e- -> 2(MnO2) + 4H2O
8. Add the two balanced half-reactions together:
Combine the two half-reactions to obtain the balanced equation:
6(CN-) + 6(OH-) + 2(MnO4-) -> 6(CNO-) + 3H2O + 6e- + 2(MnO2)
Now, the equation is balanced in a basic solution by adding hydroxide ions (OH-) to neutralize the excess hydrogen ions (H+) produced during the reaction.