1) Balance the following equation in basic conditions:
2CoCl2 + Na2O2 --> 2Co(OH)3 + 4Cl^- + 2Na^+
2) Balance the following equation in acidic conditions:
H3AsO3 + I2 --> H3AsO4 +2I^-
To balance chemical equations in basic or acidic conditions, follow these steps:
1) Write out the unbalanced equation.
2) Balance the atoms in the equation, one element at a time, starting with the most complex molecules.
3) Balance the oxygen atoms by adding water (H2O) to the appropriate side of the equation.
4) Balance the hydrogen atoms by adding hydrogen ions (H+) to the appropriate side of the equation.
5) Balance the charge on each side of the equation by adding electrons (e^-) to the appropriate side.
6) Multiply each half-reaction by an appropriate integer to balance the number of electrons transferred in each half-reaction.
7) Cancel out any duplicate species that appear on both sides of the equation.
8) Verify that the number of atoms and charges are balanced on each side of the equation.
Let's balance the given equations:
1) Balanced equation in basic conditions:
2CoCl2 + Na2O2 --> 2Co(OH)3 + 4Cl^- + 2Na^+
Step-by-step solution:
First, we balance the metals, starting with cobalt (Co). We place a coefficient of 2 in front of CoCl2 and Co(OH)3:
2CoCl2 + Na2O2 --> 2Co(OH)3 + 4Cl^- + 2Na^+
Next, we balance the non-metals by adding coefficients to Na2O2:
2CoCl2 + 2Na2O2 --> 2Co(OH)3 + 4Cl^- + 2Na^+
Now we have 4 oxygen (O) atoms only on the right side, so we need to balance them by adding water (H2O):
2CoCl2 + 2Na2O2 --> 2Co(OH)3 + 4Cl^- + 2Na^+ + 2H2O
Finally, we check the charges and atoms on each side of the equation. We can see that the charges and atoms are balanced, and there are no duplicate species, so the equation is now balanced in basic conditions:
2CoCl2 + 2Na2O2 --> 2Co(OH)3 + 4Cl^- + 2Na^+ + 2H2O
2) Balanced equation in acidic conditions:
H3AsO3 + I2 --> H3AsO4 + 2I^-
Step-by-step solution:
First, we balance the iodine (I) atoms by placing a coefficient of 2 in front of iodine on the left side:
H3AsO3 + 2I2 --> H3AsO4 + 2I^-
Next, we balance the oxygen (O) atoms by adding water (H2O) to the appropriate side of the equation:
H3AsO3 + 2I2 --> H3AsO4 + 6H2O + 2I^-
Now, we balance the hydrogen (H) atoms by adding H+ ions:
H3AsO3 + 2I2 + 6H+ --> H3AsO4 + 6H2O + 2I^-
Next, we balance the charges by adding electrons (e^-) to the appropriate side:
H3AsO3 + 2I2 + 6H+ + 8e^- --> H3AsO4 + 6H2O + 2I^-
Now, we multiply each half-reaction by an appropriate integer to balance the number of electrons transferred. In this case, we need to multiply the reduction half-reaction (left side) by 4 and the oxidation half-reaction (right side) by 2:
4H3AsO3 + 8I2 + 24H+ + 32e^- --> 4H3AsO4 + 24H2O + 8I^-
Finally, we cancel out any duplicate species and verify that the charges and atoms are balanced on both sides. We can see that the charges and atoms are balanced, and there are no duplicate species, so the equation is now balanced in acidic conditions:
4H3AsO3 + 8I2 + 24H+ + 32e^- --> 4H3AsO4 + 24H2O + 8I^-