balance half equations in basic solution
ClO3- to ClO4-
Zn to zn2+
BrO3- to Br-
IO3- to I2
Sn2+ to Sn4+
Here are the rules.
http://www.chemteam.info/Redox/Redox.html
The first one as an example of how these are applied.
ClO3^- ==> ClO4^-
1. Oxidation state Cl on left is +5; on the right is +7. Add electrons to the appropriate side to balance the change in oxdn state.
ClO3^- ==> ClO4^- + 2e
2. Count the charge on the left and right and balance by adding either
a)H+ in acid solution or
b)OH^- in basic solution.
I see -1 charge on left and -3 on right; therefore,
ClO3^- + 2OH^- ==> ClO4^- + 2e
3. Now add H2O to the appropriate side to balance H (and O).
ClO3^- + 2OH^- ==> ClO4^- + 2e + H2O
4. check it for
a. atoms balance
b. charge balance
c. change in oxidation state balance.
If you have any questions about this or the others, please post your work and explain in detail exactly what trouble you are having. These are simple enough if you follow the rules. By the way, your question would have been answered quicker if you had not changed your screen name for your posts. It helps us help you if you use the same screen name.
Why did the half equations go to therapy? Because they needed to find balance in their lives! Let's help them out:
1. ClO3- to ClO4-:
First, let's balance the oxygen by adding H2O. This gives us:
ClO3- → ClO4- + H2O
Next, we balance the hydrogen by adding H+ ions. The final balanced equation is:
ClO3- + H2O → ClO4- + H+
2. Zn to Zn2+:
Zinc (Zn) is losing two electrons to become Zn2+. So the balanced equation is:
Zn → Zn2+ + 2e-
3. BrO3- to Br-:
Let's start by balancing the oxygen atoms by adding H2O:
BrO3- + 6H2O → 6OH- + Br-
Now, we balance the hydrogen atoms by adding H+ ions:
BrO3- + 6H2O + 6H+ → 6OH- + Br-
Finally, we balance the charge by adding electrons:
8BrO3- + 9H2O + 6H+ + 6e- → 6Br- + 18OH-
4. IO3- to I2:
To balance the iodine atoms, we need to add I- ions as reactants:
IO3- + I- → I2
Next, we balance the oxygen atoms by adding H2O:
IO3- + I- + 6H2O → I2 + 6OH-
Finally, we balance the charge by adding electrons:
IO3- + I- + 6H2O + 6e- → I2 + 6OH-
5. Sn2+ to Sn4+:
Tin (Sn) is losing two electrons to become Sn4+. So the balanced equation is:
Sn2+ → Sn4+ + 2e-
There you have it! These equations are now balanced and ready to put on a great show!
To balance half equations in a basic solution, follow these steps:
1. Identify the oxidation and reduction half-reactions.
2. Balance the number of atoms on each side of the equation, excluding oxygen and hydrogen atoms.
3. Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen.
4. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen.
5. Balance the charge by adding electrons (e-) to one side of the equation.
6. Multiply the half-reactions by the appropriate number to make the number of electrons equal in both half-reactions.
7. Add the half-reactions together and cancel out any identical species on both sides of the equation.
8. If necessary, add OH- ions to both sides of the equation to neutralize any remaining H+ ions and form water (H2O).
9. Check that the atoms, charges, and elements are balanced on both sides of the equation.
Let's go through each half-reaction:
1. ClO3- to ClO4-:
Half-reaction:
ClO3- → ClO4-
To balance this equation, in basic solution, follow these steps:
1. Balance the oxygen atoms by adding water molecules:
ClO3- → ClO4- + H2O
2. Balance the hydrogen atoms by adding hydrogen ions (H+):
ClO3- + 2H+ → ClO4- + H2O
3. Balance the charge by adding electrons (e-):
ClO3- + 2H+ + 2e- → ClO4- + H2O
2. Zn to Zn2+:
Half-reaction:
Zn → Zn2+
To balance this equation, in basic solution, follow these steps:
1. Balance the charge by adding electrons (e-):
Zn → Zn2+ + 2e-
3. BrO3- to Br-:
Half-reaction:
BrO3- → Br-
To balance this equation, in basic solution, follow these steps:
1. Balance the oxygen atoms by adding water molecules:
BrO3- → Br- + 3H2O
2. Balance the hydrogen atoms by adding hydrogen ions (H+):
BrO3- + 6H+ → Br- + 3H2O
3. Balance the charge by adding electrons (e-):
BrO3- + 6H+ + 6e- → Br- + 3H2O
4. IO3- to I2:
Half-reaction:
IO3- → I2
To balance this equation, in basic solution, follow these steps:
1. Balance the oxygen atoms by adding water molecules:
IO3- → I2 + 3H2O
2. Balance the hydrogen atoms by adding hydrogen ions (H+):
IO3- + 6H+ → I2 + 3H2O
3. Balance the charge by adding electrons (e-):
IO3- + 6H+ + 6e- → I2 + 3H2O
5. Sn2+ to Sn4+:
Half-reaction:
Sn2+ → Sn4+
To balance this equation, in basic solution, follow these steps:
1. Balance the charge by adding electrons (e-):
Sn2+ → Sn4+ + 2e-
These are the balanced half-equations in basic solution:
1. ClO3- to ClO4-:
ClO3- + 2H+ + 2e- → ClO4- + H2O
2. Zn to Zn2+:
Zn → Zn2+ + 2e-
3. BrO3- to Br-:
BrO3- + 6H+ + 6e- → Br- + 3H2O
4. IO3- to I2:
IO3- + 6H+ + 6e- → I2 + 3H2O
5. Sn2+ to Sn4+:
Sn2+ → Sn4+ + 2e-
To balance half-equations in basic solution, you need to follow the following steps:
1. Write down the half-equations:
ClO3- → ClO4-
Zn → Zn2+
BrO3- → Br-
IO3- → I2
Sn2+ → Sn4+
2. Balance the atoms other than oxygen and hydrogen:
ClO3- → ClO4-
To balance the number of chlorine atoms, add 1 Cl- on the right side:
ClO3- → ClO4- + Cl-
Zn → Zn2+
No need to balance additional atoms as they are already balanced.
BrO3- → Br-
To balance the number of bromine atoms, add 3 Br- on the right side:
BrO3- + 3 Br- → Br-
IO3- → I2
To balance the number of iodine atoms, add 3 I2 on the right side:
IO3- + 3 I2 → I2
Sn2+ → Sn4+
No need to balance additional atoms as they are already balanced.
3. Balance oxygen atoms by adding water molecules (H2O) to the side that needs oxygen:
ClO3- → ClO4- + Cl- + 2 H2O
Zn → Zn2+
No oxygen atoms to balance.
BrO3- + 3 Br- → Br- + 3 H2O
IO3- + 3 I2 → I2 + 6 H2O
Sn2+ → Sn4+
No oxygen atoms to balance.
4. Balance hydrogen atoms by adding hydrogen ions (H+) to the side that needs hydrogen:
ClO3- → ClO4- + Cl- + 2 H2O + 4 H+
Zn → Zn2+
No hydrogen atoms to balance.
BrO3- + 3 Br- + 6 H+ → Br- + 3 H2O
IO3- + 3 I2 + 6 H+ → I2 + 6 H2O
Sn2+ → Sn4+
No hydrogen atoms to balance.
5. Balance the charges by adding electrons (e-):
ClO3- + 6 e- → ClO4- + Cl- + 2 H2O + 4 H+
Zn → Zn2+ + 2 e-
BrO3- + 3 Br- + 6 H+ + 6 e- → Br- + 3 H2O
IO3- + 3 I2 + 6 H+ + 6 e- → I2 + 6 H2O
Sn2+ → Sn4+ + 2 e-
Now the half-equations are balanced in basic solution. The number of electrons (e-) should be the same in both half-equations.