Balance the following equation in acidic conditions.
S2O3^2- +Cu ^2+ > S4O6^2- +Cu^+
2S2O3^2- +2Cu ^2+ > S4O6^2- + 2Cu^+
Oxidation-Reduction Balancing
Acidic Solution
Cr2O7 + C2H4O -> C2H4O2 + Cr3+
To balance the given chemical equation in acidic conditions, we need to follow some steps:
Step 1: Count the number of atoms of each element on both sides of the equation.
On the left side:
Sulfur (S): 2
Oxygen (O): 9
Copper (Cu): 1
On the right side:
Sulfur (S): 5
Oxygen (O): 12
Copper (Cu): 2
Step 2: Balance the atoms that are not in balance starting with those that appear in the least number.
Let's balance the sulfur atoms first:
S2O3^2- +Cu^2+ > 2S4O6^2- +Cu^+
Now, the sulfur atoms are balanced.
Step 3: Balance the oxygen atoms using water (H2O).
On the left side:
Oxygen (O): 9
On the right side:
Oxygen (O): 24
We need to add 15 H2O molecules to the left side to balance the oxygen atoms:
S2O3^2- +Cu^2+ + 15H2O > 2S4O6^2- +Cu^+
Step 4: Balance the hydrogen atoms using H+ ions.
On the left side:
Hydrogen (H): 30
On the right side:
Hydrogen (H): 30
The hydrogen atoms are already balanced.
Step 5: Finally, balance the charge by adding electrons (e-) to one side of the equation.
On the left side, the total charge is 4- (for each S2O3^2- ion):
So, we need 4 electrons on the left side:
S2O3^2- +Cu^2+ + 15H2O + 4e- > 2S4O6^2- +Cu^+
The balanced chemical equation under acidic conditions is:
S2O3^2- +Cu^2+ + 15H2O + 4e- > 2S4O6^2- +Cu^+
To check the balance, count the atoms of each element and the charges on both sides of the equation.