Balance the equation using ion-electron method
(S2O3)2- + (OCl)- = (SO4)2- + Cl-
I will do each individually. You can then put them together.
2H^+ + OCl^- + 2e ==> Cl^- + H2O
(S2O3)^2- + 5H2O ==> 2(SO4)^2- + 8e + 10H^+
Multiply equation 1 by 4 and add to equation 2 for the final equation.
To balance the given equation using the ion-electron method, we need to ensure that the number of atoms on both sides of the equation are equal.
Given:
(S2O3)2- + (OCl)- = (SO4)2- + Cl-
Let's start by examining the elements involved:
Sulfur (S)
Oxygen (O)
Chlorine (Cl)
We can split the equation into half-reactions, one for the reduction and one for the oxidation:
Oxidation half-reaction: (S2O3)2- → (SO4)2-
Reduction half-reaction: (OCl)- → Cl-
In the oxidation half-reaction, the Sulfur (S) atom gets oxidized, going from a -2 oxidation state to +6 in (SO4)2-. To balance the sulfur atoms, we need to add 3 electrons to the left side:
(S2O3)2- → (SO4)2- + 3e-
In the reduction half-reaction, the Chlorine (Cl) atom gets reduced, going from a +1 oxidation state to a -1 charge in Cl-. To balance the chlorine atoms and charges, we need to add one electron to the right side:
(OCl)- + e- → Cl-
Now, we need to balance the oxygen atoms by adding water (H2O) molecules and hydrogen ions (H+):
In the oxidation half-reaction, there are 3 oxygen atoms on the right side but only 2 on the left side. Adding 1.5 water molecules on the left side balances the oxygen atoms:
(S2O3)2- + 1.5H2O → (SO4)2- + 3e-
In the reduction half-reaction, there is 1 oxygen atom on the right side but none on the left side. Adding 0.5 water molecules on the left side balances the oxygen atoms:
(OCl)- + e- + 0.5H2O → Cl-
Now, let's balance the charges by adding electrons to the right side of the oxidation half-reaction:
(S2O3)2- + 1.5H2O → (SO4)2- + 3e- + 6H+
To balance the charges in the reduction half-reaction, we can multiply the entire equation by 3:
3(OCl)- + 3e- + 1.5H2O → 3Cl- + 3e-
Combining the two half-reactions, we cancel out the electrons:
6(OCl)- + 6e- + 3H2O + (S2O3)2- → 6Cl- + 3H2O + (SO4)2- + 3e-
Now we can remove the water that appears on both sides:
6(OCl)- + (S2O3)2- → 6Cl- + (SO4)2-
Finally, we can simplify the equation by removing the spectator ions that appear on both sides:
6OCl- + 3S2O3 2- → 6Cl- + 3SO4 2-
And there you have it. The balanced equation using the ion-electron method is:
6OCl- + 3S2O3 2- → 6Cl- + 3SO4 2-
To balance the given equation using the ion-electron method, follow these steps:
Step 1: Write the unbalanced equation.
(S2O3)2- + (OCl)- = (SO4)2- + Cl-
Step 2: Separate the equation into half-reactions. Identify which species undergo oxidation and which undergo reduction.
Half-reaction 1: (S2O3)2- -> (SO4)2-
Half-reaction 2: (OCl)- -> Cl-
Step 3: Balance the elements individually.
Let's start with the sulfur (S) atoms:
In Half-reaction 1, there are two sulfur atoms on both sides of the equation.
In Half-reaction 2, there is one sulfur atom on the left side and no sulfur atom on the right side.
To balance the sulfur atoms, add a coefficient of 2 in front of the OCl- in Half-reaction 2:
(S2O3)2- -> (SO4)2-
2(OCl)- -> 2Cl-
Now, the sulfur atoms are balanced.
Step 4: Balance the oxygen atoms by adding water (H2O) molecules.
In Half-reaction 1, there are three oxygen atoms on the right side (SO42-), while there are none on the left side.
Add 3 H2O molecules to the left side:
(S2O3)2- + 3H2O -> (SO4)2-
In Half-reaction 2, there are no oxygen atoms on either side, so no water molecules need to be added.
Step 5: Balance the hydrogen (H) atoms by adding hydrogen ions (H+).
In Half-reaction 1, there are six hydrogen atoms on the left side (3 H2O), while there are none on the right side.
Add 6 H+ ions to the right side:
(S2O3)2- + 3H2O -> (SO4)2- + 6H+
In Half-reaction 2, there are no hydrogen atoms on either side, so no hydrogen ions need to be added.
Step 6: Balance the charges by adding electrons (e-).
In Half-reaction 1, the overall charge is 2- on the left side (S2O32-), while the right side has a 2- charge (SO42-).
To balance the charges, add 2 electrons (e-) to the left side:
(S2O3)2- + 3H2O + 2e- -> (SO4)2- + 6H+
In Half-reaction 2, the overall charge is 1- on the left side (OCl-), while the right side has a 1- charge (Cl-).
There is no need to add electrons in this half-reaction.
Step 7: Add both half-reactions together and cancel out common terms.
(S2O3)2- + 3H2O + 2e- -> (SO4)2- + 6H+ ... (1)
2(OCl)- -> 2Cl- ... (2)
By multiplying equation (2) by 3 and adding it to equation (1), the electrons will cancel out.
Multiply equation (2) by 3:
6(OCl)- -> 6Cl-
Add equation (1) and the multiplied equation (2):
(S2O3)2- + 3H2O + 2e- + 6(OCl)- -> (SO4)2- + 6H+ + 6Cl-
Step 8: Simplify the equation and confirm that it is balanced.
(S2O3)2- + 3H2O + 6OCl- -> (SO4)2- + 6H+ + 6Cl-
Now, the equation is balanced using the ion-electron method.