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Question: Balance the given equation: \({H_2}{O_2} + {I_2} = HI{O_3} + {H_2}O\)...

Balance the given equation: H2O2+I2=HIO3+H2O{H_2}{O_2} + {I_2} = HI{O_3} + {H_2}O

Explanation

Solution

One approach to adjust redox responses is by monitoring the electron move utilizing the oxidation quantities of every one of the iotas/atoms. For the oxidation-number-change technique, start with the unbalanced skeleton condition.

Complete step by step answer:
We will try to solve this question on the basis of balancing the redox reactions. In this process, we will assign oxidation number to the atoms of those elements which undergo ON change in the redox reaction and then we will try to balance the above reaction.

The reaction that is given to us is:
H2O2+I2=HIO3+H2O{H_2}{O_2} + {I_2} = HI{O_3} + {H_2}O
We were given that the change in ON is 10. Now let us try to balance the equation:
First, let us try to calculate the number of hydrogens on both the sides of the equation, considering the Left Hand Side (LHS) of the reaction:
H2O2+I2{H_2}{O_2} + {I_2}; here the number of hydrogens is;
H2O2{H_2}{O_2} - 1×21 \times 2= 2
Now let us consider Right Hand Side (RHS) of the reaction:
HIO3+H2OHI{O_3} + {H_2}O; here the number of hydrogens is;
(HIO3HI{O_3} = 1×11 \times 1) + (H2O{H_2}O= 2×12 \times 1 )
1+2=3\Rightarrow 1 + 2 = 3

Now, let us calculate the change in charge in the reaction:
If we carefully calculate and observe, we find that from H2O22H2O{H_2}{O_2} \to 2{H_2}O, the change in charge is 1×22×2 - 1 \times 2 \to - 2 \times 2
And, when we calculate the same for I22HIO3{I_2} \to 2HI{O_3}, we find that the change is
0+50 \to + 5
Keeping this change in mind we need to balance the equation as:
5H2O2+I22HIO3+4H2O5{H_2}{O_2} + {I_2} \to 2HI{O_3} + 4{H_2}O is the required balanced equation we require.

The rules for balancing redox reactions using oxidation number rule:
1.Write the condition, show the oxidation number of every component and distinguish the component which is going through change in oxidation number. Identify the oxidizing just as a reducing agent.
2.Multiply the equation of the oxidizing and reducing agent by suitable integers to adjust the absolute increment or diminishes in oxidation number. Balance all atoms other than HH and OO, at long last balance HH and OO by adding water atoms.

Note:
The oxidation number technique, monitors electrons picked up when a substance is diminished and the electrons lost when a substance is oxidized. Dole out oxidation numbers to every one of the particles in the condition and compose the numbers over the iota/atom.