Balance the following redox equations in basic medium by oxidation number method: a) $$Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ }Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}$$b) $$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}\text{ }\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+C{{l}^{-}}_{\left( aq \right)}$$ c)$$~M{{n}^{2+}}_{\left( aq \right)}\text{ }+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}\to Mn{{O}_{2}}_{\left( s \right)}+{{H}_{2}}{{O}_{\left( l \right)}}$$d)$$Cu{{\left( OH \right)}_{2}}_{\left( s \right)}\text{ }+\text{ }{{N}_{2}}{{H}_{4}}_{\left( aq \right)}\text{ }\to \text{ }C{{u}_{\left( s \right)}}+{{N}_{2}}_{\left( g \right)}$$

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Vedantu Content Team · Accepted Answer

Hint: Redox reactions are those in which oxidation and reduction occurs simultaneously. In oxidation the element gains electrons and in reduction the element loses electrons. Also, the oxidation number increases when the element is oxidised and it decreases when the element is reduced.Complete answer: For each reaction, you have to find the changed oxidation number of each element. In order to balance the reaction, water molecules ${{H}_{2}}O$, hydrogen ions $\left( H+ \right)$and Hydroxide ions $\left( OH- \right)$are added on either side. Let us now see each reaction individually,a) $$Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ }Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}$$ The unbalanced redox equation is $$Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ }Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}$$Here, Zn and N are balanced and O and H are not.The oxidation number of Zn changes from 0 to +6. The change in oxidation number is by +6.The oxidation number of N changes from +5 to -3. The change in oxidation number is by 8.The increase in oxidation number is balanced out by decrease in oxidation number by multiplying $$Zn\text{ and }Zn{{\left( OH \right)}_{6}}^{-2}$$by 8.$$8Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ 8}Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}$$The O atoms will be balanced by adding 3 water molecules to the RHS of the equation.$$8Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ 8}Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}+\text{ 3}{{\text{H}}_{2}}O$$The hydrogen atoms are balanced by adding 9H+ to the LHS of the equation.$$8Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}+\text{ 9H+}\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ 8}Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}+\text{ 3}{{\text{H}}_{2}}O$$The reaction occurs in the basic medium, so $9\text{ }OH-$ ions will be added on both sides of the equation,$$8Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}+\text{ 9}\left( \text{H+} \right)\text{ +9 OH- }\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ 8}Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}+\text{ 3}{{\text{H}}_{2}}O\text{ + 9OH-}$$The 9 H+ and OH- atoms on the LHS combine to form 9 ${{H}_{2}}O$ molecules out of which 3 get cancelled from the RHS, to get 6${{H}_{2}}O$ molecules.$$8Z{{n}_{\left( s \right)}}+\text{ }N{{O}_{3}}_{\left( aq \right)}+6{{\text{H}}_{2}}O\text{ }\to \text{ }N{{H}_{3}}_{\left( aq \right)}+\text{ 8}Zn{{\left( OH \right)}_{6}}{{^{-2}}_{\left( aq \right)}}+\text{ 9OH-}$$This is the balanced chemical equation by oxidation method.b) $$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}\text{ }\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+C{{l}^{-}}_{\left( aq \right)}$$Here, Mn is balanced and Cl, O and H are not.The oxidation number of Mn changes from +2 to +3. The change in oxidation number is by +1.The oxidation number of Cl changes from -2 to -1. The change in oxidation number is by -1.The increase in oxidation number is balanced out by a decrease in oxidation number by multiplying $C{{l}^{-}}$by 2.$$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}\text{ }\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+2C{{l}^{-}}_{\left( aq \right)}$$The O atoms will be balanced by adding 2 water molecules to the RHS of the equation.$$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}\text{ }\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+2C{{l}^{-}}_{\left( aq \right)}+2{{H}_{2}}O$$The hydrogen atoms are balanced by adding 3H+ to the LHS of the equation.$$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}\text{+ 3H+}\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+2C{{l}^{-}}_{\left( aq \right)}+2{{H}_{2}}O$$The reaction occurs in basic medium, so $\text{3 }OH-$ ions will be added on both sides of the equation,$$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}\text{+3}\left( OH- \right)\text{+ 3H+}\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+2C{{l}^{-}}_{\left( aq \right)}+2{{H}_{2}}O+\text{ 3}\left( OH- \right)$$The 3 H+ and OH- atoms on the LHS combine to form 3${{H}_{2}}O$ molecules out of which 2 get cancelled from the RHS, to get 1${{H}_{2}}O$ molecule.$$MnC{{l}_{2}}_{\left( aq \right)}+\text{ }H{{O}_{2}}{{^{-}}_{\left( aq \right)}}+1{{H}_{2}}O\to Mn{{\left( OH \right)}_{3}}_{\left( s \right)}+2C{{l}^{-}}_{\left( aq \right)}$$This is the balanced chemical equation by oxidation method.c)$$~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}\to \text{ Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}$$The unbalanced redox equation is $$~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}\to \text{ Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}$$Here, H is balanced and O and Mn are not balanced.The oxidation number of Zn changes from +2 to +4. The change in oxidation number is by +2.The increase in oxidation number is balanced out by decrease in oxidation number by multiplying $$Mn\text{ and Mn}{{\text{O}}_{2}}$$ by 2.$$2~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}\to \text{ 2Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}$$The O atoms will be balanced by adding 3 water molecules to the LHS of the equation.$$2~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}+\text{ 3 }{{\text{H}}_{2}}O\to \text{ 2Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}$$The hydrogen atoms are balanced by adding 6H+ to the RHS of the equation.$$2~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}+\text{ 3 }{{\text{H}}_{2}}O\to \text{ 2Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}+\text{ 6H+}$$The reaction occurs in basic medium, so $\text{6 }OH-$ ions will be added on both sides of the equation,$$2~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}+\text{ 3 }{{\text{H}}_{2}}O+\text{ 6 OH-}\to \text{ 2Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}+\text{ 6}\left( \text{H+} \right)\text{ + 6 OH-}$$The 6 H+ and OH- atoms on the RHS combine to form 6 ${{H}_{2}}O$ molecules out of which 3 get cancelled from the LHS, to get 3${{H}_{2}}O$ molecules on the RHS.$$2~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}+\text{ 6 OH-}\to \text{ 2Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ }{{\text{H}}_{2}}{{O}_{\left( l \right)}}+\text{ 3 }{{\text{H}}_{2}}O$$On the LHS there will be total 4 water molecules.$$2~M{{n}^{+2}}_{\left( aq \right)}+\text{ }{{H}_{2}}{{O}_{2}}_{\left( aq \right)}+\text{ 6 OH-}\to \text{ 2Mn}{{\text{O}}_{2}}_{\left( s \right)}+\text{ 4}{{\text{H}}_{2}}{{O}_{\left( l \right)}}$$This is the balanced chemical equation by oxidation method.d) $$Cu{{\left( OH \right)}_{2}}_{\left( s \right)}\text{ }+\text{ }{{N}_{2}}{{H}_{4}}_{\left( aq \right)}\text{ }\to \text{ }C{{u}_{\left( s \right)}}+{{N}_{2}}_{\left( g \right)}$$The unbalanced redox equation is $$Cu{{\left( OH \right)}_{2}}_{\left( s \right)}\text{ }+\text{ }{{N}_{2}}{{H}_{4}}_{\left( aq \right)}\text{ }\to \text{ }C{{u}_{\left( s \right)}}+{{N}_{2}}_{\left( g \right)}$$Here, Cu and N are balanced and O and H are not.The oxidation number of Cu changes from +2 to 0. The change in oxidation number is by 2.The oxidation number of N changes from -2to +5. The change in oxidation number is by 7.The hydrogen atoms are balanced by adding 4H+ to the LHS of the equation.$$Cu{{\left( OH \right)}_{2}}_{\left( s \right)}\text{ }+\text{ }{{N}_{2}}{{H}_{4}}_{\left( aq \right)}\text{ }\to \text{ }C{{u}_{\left( s \right)}}+{{N}_{2}}_{\left( g \right)}+4H+$$The reaction occurs in basic medium, so $\text{4 }OH-$ ions will be added on both sides of the equation,$$Cu{{\left( OH \right)}_{2}}_{\left( s \right)}\text{ }+\text{ }{{N}_{2}}{{H}_{4}}_{\left( aq \right)}+\text{ 4OH-}\to \text{ }C{{u}_{\left( s \right)}}+{{N}_{2}}_{\left( g \right)}+4H+4OH-$$The 4 H+ and OH- atoms on the RHS combine to form 4${{H}_{2}}O$molecules $$Cu{{\left( OH \right)}_{2}}_{\left( s \right)}\text{ }+\text{ }{{N}_{2}}{{H}_{4}}_{\left( aq \right)}+\text{ 4OH-}\to \text{ }C{{u}_{\left( s \right)}}+{{N}_{2}}_{\left( g \right)}+4{{H}_{2}}O$$This is the balanced chemical equation by oxidation method.Note:The oxidation number is basically the count of electrons that atoms in a molecule can share, lose or gain while forming chemical bonds with other atoms of a different element. Oxidation number is also referred to as oxidation state.

Balance the following redox equations in basic medium by oxidation number method: a) Zn(s)+ NO3(aq)→ NH3(aq)+ Zn(OH)6−2(aq)b) MnCl2(aq)+ HO2−(aq) →Mn(OH)3(s)+Cl−(aq) c) Mn2+(aq) + H2O2(aq)→MnO2(s)+H2O(l)d)Cu(OH)2(s) + N2H4(aq) → Cu(s)+N2(g)

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