Question
Describe the steps involved in balancing redox reactions by ion electron method $($Half reaction method$).$
✓
Answer
In this method two half equations are balanced separately and then added together to give balanced equation. Following steps are involved.
Step $1:$ Write unbalanced equation for the redox reaction, assign oxidation number to all the atoms in the reactants and products. Divide the equation into two half equations. One half equation involves increase in oxidation number and another involves decrease in oxidation number $($Write two half equations separately$).$
Step $2:$ Balance the atoms except $O$ and $H$ in each half equation. Balance oxygen atom by adding $H_2O$ to the side with less $O$ atoms.
Step $3:$ Balance $H$ atoms by adding $H^+$ ions to the side having less $H$ atoms.
Step $4:$ Balance the charges by adding appropriate number of electrons to the right side of oxidation half equation and to the left of reduction half equation.
Step $5:$ Multiply half equation by suitable factors to equalize the number of electrons in two half equations. Add two half equations and cancel the number of electrons on both sides of equation.
Step $6:$ If the reaction occurs in basic medium then add $OH^–$ ions, equal to number of $H^+$ ions on both sides of equation. The $H^+$ and $OH^–$ ions on same side of equation combine to give $H_2O$ molecules.
Check that the equation is balanced in both, the atoms and the charges.
Step $1:$ Write unbalanced equation for the redox reaction, assign oxidation number to all the atoms in the reactants and products. Divide the equation into two half equations. One half equation involves increase in oxidation number and another involves decrease in oxidation number $($Write two half equations separately$).$
Step $2:$ Balance the atoms except $O$ and $H$ in each half equation. Balance oxygen atom by adding $H_2O$ to the side with less $O$ atoms.
Step $3:$ Balance $H$ atoms by adding $H^+$ ions to the side having less $H$ atoms.
Step $4:$ Balance the charges by adding appropriate number of electrons to the right side of oxidation half equation and to the left of reduction half equation.
Step $5:$ Multiply half equation by suitable factors to equalize the number of electrons in two half equations. Add two half equations and cancel the number of electrons on both sides of equation.
Step $6:$ If the reaction occurs in basic medium then add $OH^–$ ions, equal to number of $H^+$ ions on both sides of equation. The $H^+$ and $OH^–$ ions on same side of equation combine to give $H_2O$ molecules.
Check that the equation is balanced in both, the atoms and the charges.
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