how to balance the following redox
1 ClO- +I- = Cl- +I2
also I2+S2O3(2-) == I- +S4O6(2-)
I2 + S2O3^2- ==> 2I^- + S4O6^2-
2H^+ + ClO^- + 2I^- ==> Cl^- + I2 + H2O
To balance the given redox equation:
Step 1: Assign oxidation numbers to each element in the equation.
The oxidation state of chlorine (Cl) in ClO- is -1, while in Cl-, it is 0.
The oxidation state of iodine (I) in I- is -1, and in I2, it is 0.
Step 2: Identify the oxidized and reduced species.
In this case, chlorine is reduced (its oxidation number decreases from -1 to 0), and iodine is oxidized (its oxidation number increases from -1 to 0).
Step 3: Write separate half-reactions for oxidation and reduction.
Oxidation half-reaction: I- → I2
Reduction half-reaction: ClO- → Cl-
Step 4: Balance the atoms in each half-reaction, excluding oxygen and hydrogen.
Oxidation half-reaction: 2I- → I2
Reduction half-reaction: ClO- → 2Cl-
Step 5: Balance the oxygen atoms by adding water (H2O) molecules.
Oxidation half-reaction: 2I- → I2 + 2e-
Reduction half-reaction: ClO- + H2O → 2Cl- + 2e-
Step 6: Balance the hydrogen atoms by adding hydrogen ions (H+) or hydroxide ions (OH-).
Since this reaction is taking place in a basic solution (OH- present), we will balance it using hydroxide ions (OH-).
Oxidation half-reaction: 2I- → I2 + 2e-
Reduction half-reaction: ClO- + H2O + 2OH- → 2Cl- + 2e- + 2OH-
Step 7: Balance the charges in each half-reaction by adding electrons (e-).
Oxidation half-reaction: 2I- → I2 + 2e-
Reduction half-reaction: ClO- + H2O + 2OH- → 2Cl- + 2e- + 2OH-
Step 8: Multiply each half-reaction by an appropriate factor to make the number of electrons equal.
To balance the electrons, we need to multiply the oxidation half-reaction by 2.
2I- → I2 + 4e-
2ClO- + 2H2O + 4OH- → 4Cl- + 4e- + 4OH-
Step 9: Combine the balanced half-reactions.
2I- + 2ClO- + 2H2O + 4OH- → I2 + 4Cl- + 4e- + 4OH-
Step 10: Simplify the equation by canceling out the common species on each side.
2I- + 2ClO- + 2H2O → I2 + 4Cl- + 4OH-
And there you have it! The balanced redox equation is:
2I- + 2ClO- + 2H2O → I2 + 4Cl- + 4OH-