Question
Balance the following redox equation by half reaction method
✓
Answer
Balance the following redox equation by half reaction method
i. $H _2 C _2 O _{4( aq )}+ MnO _{4(a q)}^{-} \rightarrow CO _{2( g )}+ Mn _{( aq )}^{2+}$
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding $4H_2O$ to the right side of reduction half equation.
Step 3: Balance $H$ atoms by adding $H ^{+}$ions to the side with less $H$. Hence, add $2 H ^{+}$ions to the right side of oxidation half equation and $8 H ^{+}$ions to the left side of reduction half equation.
Step 4: Now add $2$ electrons to the right side of oxidation half equation and $5$ electrons to the left side of reduction half equation to balance the charges.
Step 5: Multiply oxidation half equation by $5$ and reduction half equation by $2$ to equalize number of electrons in two half equations. Then add two half equation.
ii. $Bi ( OH )_{3( s )}+ SnO _{2( aq )}^{2-} \longrightarrow SnO _{3( aq )}^{2-}+ Bi _{( s )}$
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
Step 2: Balance half equations for O atoms by adding $H_2O$ to the side with less O atoms. Add $1H_2O$ to left side of oxidation half equation and $3H_2O$ to the right side of reduction half equation.
Step 3: Balance H atoms by adding $H^+$ ions to the side with less H. Hence, add $2H^+$ ions to the right side of oxidation half equation and $3H^+$ ions to the left side of reduction half equation.
Step 4: Now add 2 electrons to the right side of oxidation half equation and $3$ electrons to the left side of reduction half equation to balance the charges.
Step 5: Multiply oxidation half equation by $3$ reduction half equation by $2$ to equalize number of electrons in two half equations. Then add two half equation.
Reaction occurs in basic medium. However, $H^+$ ions cancel out and the reaction is balanced. Hence, no need to add $OH^–$ ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:
i. $H _2 C _2 O _{4( aq )}+ MnO _{4(a q)}^{-} \rightarrow CO _{2( g )}+ Mn _{( aq )}^{2+}$
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding $4H_2O$ to the right side of reduction half equation.
Step 3: Balance $H$ atoms by adding $H ^{+}$ions to the side with less $H$. Hence, add $2 H ^{+}$ions to the right side of oxidation half equation and $8 H ^{+}$ions to the left side of reduction half equation.
Step 4: Now add $2$ electrons to the right side of oxidation half equation and $5$ electrons to the left side of reduction half equation to balance the charges.
Step 5: Multiply oxidation half equation by $5$ and reduction half equation by $2$ to equalize number of electrons in two half equations. Then add two half equation.
ii. $Bi ( OH )_{3( s )}+ SnO _{2( aq )}^{2-} \longrightarrow SnO _{3( aq )}^{2-}+ Bi _{( s )}$
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.
Step 2: Balance half equations for O atoms by adding $H_2O$ to the side with less O atoms. Add $1H_2O$ to left side of oxidation half equation and $3H_2O$ to the right side of reduction half equation.
Step 3: Balance H atoms by adding $H^+$ ions to the side with less H. Hence, add $2H^+$ ions to the right side of oxidation half equation and $3H^+$ ions to the left side of reduction half equation.
Step 4: Now add 2 electrons to the right side of oxidation half equation and $3$ electrons to the left side of reduction half equation to balance the charges.
Step 5: Multiply oxidation half equation by $3$ reduction half equation by $2$ to equalize number of electrons in two half equations. Then add two half equation.
Reaction occurs in basic medium. However, $H^+$ ions cancel out and the reaction is balanced. Hence, no need to add $OH^–$ ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:
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