balance equation by electron balancing method
cl2 + oh- = cl- + clO3- + h2O
You really do need to find the caps key on your keyboard and learn to use it.
Cl2 + OH^- = Cl^- + ClO3^- + H2O
Separate into two half equations.
Cl2 ==>Cl^- and Cl2 ==> ClO3^-
Balance each separately.
Cl2 + 2e--> 2Cl^- for the first one, then
Cl2 ==> 2ClO3^-. Cl2 has oxidations state of 0 and on the right it is +10 so
Cl2 --> 2ClO3^- + 10e. Now balance the charge with OH^-.
Cl2 + 12 OH^- ==> 2ClO3^- + 10e. Now balance the H.
Cl2 + 12 OH^- ==> 2ClO3^- + 10e + 6H2O for the second one.
Now multiply the first one by 5, the second one by 1, add them and the equation will balance BUT then you can divide everything by 2 and that will be the final equation you want. Post your work if you get stuck.
Cl2 ==> 2ClO3^-. Cl2 has oxidations state of 0 and on the right it is +10 so
Cl2 --> 2ClO3^- + 10e. Now balance the charge with OH^-.
How +10????
To balance the given equation by the electron balancing method, follow these step-by-step procedures:
Step 1: Write down the unbalanced equation. In this case, it is:
Cl2 + OH- = Cl- + ClO3- + H2O
Step 2: Identify the atoms that are unbalanced. In this equation, both the chlorine (Cl) and oxygen (O) atoms are unbalanced.
Step 3: Balance the atoms that appear in fewer molecules first. Let's start with the chlorine (Cl) atoms.
Count the Cl atoms on the reactant side (left): 2 Cl on the left.
Count the Cl atoms on the product side (right): 1 Cl in Cl- and 1 Cl in ClO3- on the right.
To balance the Cl atoms:
Add a coefficient of 2 in front of Cl- and ClO3- on the right side:
Cl2 + OH- = 2Cl- + ClO3- + H2O
Now, let's move on to balancing the oxygen (O) atoms.
Count the O atoms on the reactant side (left): 1 O in OH- on the left.
Count the O atoms on the product side (right): 3 O in ClO3- and 1 O in H2O on the right.
To balance the O atoms:
Add a coefficient of 3 in front of H2O on the right side:
Cl2 + OH- = 2Cl- + ClO3- + 3H2O
Step 4: Check if the charges are balanced. In this case, the charges are already balanced, so no further adjustments are needed.
The balanced equation is:
Cl2 + OH- = 2Cl- + ClO3- + 3H2O
To balance this equation using the electron balancing method, we need to ensure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.
Let's break down the equation into two half-reactions.
Oxidation Half-Reaction:
Cl2 -> Cl- + ClO3-
Reduction Half-Reaction:
OH- -> H2O
Now, let's balance the oxidation half-reaction:
1. Start by balancing the chlorine atoms. Since there are two chlorine atoms on the left side, we need to add a coefficient of 2 in front of Cl-:
Cl2 -> 2Cl- + ClO3-
2. Next, balance the oxygen atoms. There are three oxygen atoms in ClO3-, so we need to add a coefficient of 3/2 in front of ClO3- to balance the oxygen atoms:
Cl2 -> 2Cl- + 3/2 ClO3-
Now, let's balance the reduction half-reaction:
1. Start by balancing the hydrogen atoms. There are 2 hydrogen atoms in H2O, so we need to add a coefficient of 2 in front of H2O:
OH- -> 2H2O
Now that we have balanced the two half-reactions, we need to balance the electrons to equalize the number of electrons lost and gained in the reaction.
In the oxidation half-reaction, each Cl2 molecule loses 2 electrons. Therefore, we add the coefficient 2 in front of Cl2 to reflect this:
2Cl2 -> 4Cl- + 3/2 ClO3-
In the reduction half-reaction, no electrons are either gained or lost. So, there is no need to balance electrons in this half-reaction.
To ensure the number of electrons lost and gained is equal, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2:
6Cl2 -> 12Cl- + 9/2 ClO3-
6OH- -> 12H2O
Now, to combine the two half-reactions and cancel out the common species on both sides of the equation:
6Cl2 + 6OH- -> 12Cl- + 9/2 ClO3- + 12H2O
Finally, to get rid of the fractional coefficients, we multiply the entire equation by 2:
12Cl2 + 12OH- -> 24Cl- + 9ClO3- + 24H2O
And that's how we balance the equation using the electron balancing method.