Balance the following ionic equations

(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O
(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O(iii) MnO^{-}_{4} + SO^{2-}_{3} + H^{+}\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O
(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O

 

Answers (1)

(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O

As step (i), we will be generating the unbalanced skeleton i.e.,

(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

 

Oxidation is 2I^{-}\rightarrow I_{2}+2e^{-}and Reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}


As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

 

Oxidation is 2I^{-}\rightarrow I_{2}+2e^{-}  and Reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}


Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

 

The oxidation is 2I^{-}\rightarrow I_{2}+2e^{-} and reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}

 

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

 

Hence, oxidation is  2I^{-}\rightarrow I_{2}+2e^{-}and

 

reduction is  Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

 

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction. We will now multiply the oxidation reaction by 3, which will give us,

 

The oxidation as 6I^{-}\rightarrow 3I_{2}+6e^{-} and

the reduction as  Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O


Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

 

6I^{-}+Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O+3I_{2}+6e^{-}


Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

6I^{-}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O+3I_{2}


Now, we know that the charge on the LHS is +6 which is same as the RHS of the equation, thereby enduring that the reaction is balanced.


(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O

 

As step (i), we will be generating the unbalanced skeleton i.e.,
 

Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O


 

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

 

Thus, oxidation is  Fe^{2+}\rightarrow Fe^{3+}+e^{-} and the reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}

 

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

 

Thus, oxidation is  Fe^{2+}\rightarrow Fe^{3+}+e^{-} and the reduction is Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}


Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

 

Thus, oxidation Fe^{2+}\rightarrow Fe^{3+}+e^{-}is   and

reduction is  Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}

 

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

 

Thus, oxidation isFe^{2+}\rightarrow Fe^{3+}+e^{-} and

reduction is Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

 

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

 

Thus, oxidation is6Fe^{2+}\rightarrow 6Fe^{3+}+6e^{-} and

reduction is  Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O

 

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
 

6Fe^{2+}+Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 6Fe^{3+}+6e^{-}+2Cr^{3+}+7H_{2}O

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
 

6Fe^{2+}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_{2}O

We will now verify whether all the charges are balanced.

 

So, the LHS = 6\times +2 + 1\times -2 + 14 = 24 and the RHS = 6\times +3 + 2\times +3 + 7\times 0 = 24

 

Thus the sum total is same on both the sides, therefore the solve reaction is right.


 

 (iii) MnO^{-}_{4} + SO^{2-}_{3} + H+\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O

 

As step (i), we will be generating the unbalanced skeleton i.e.,

 

MnO^{-}_{4} + SO^{2-}_{3} + H+\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O


Here in Mn undergoes reduction and S undergoes oxidation.

 

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

 

Thus, oxidation is  SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}


 

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples. But as we can see that the atoms are balanced, so the reaction will be the same as the previous step.


Thus, oxidation is SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-} and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}


 

Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.


Thus, oxidation is SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}+2H^{+} and

reduction is  MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}


 

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

 

Thus, oxidation is SO^{2-}_{3}+H_{2}O \rightarrow SO^{2-}_{4}+2e^{-}+2H^{+}and

reduction is  MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}+4H_{2}O


 

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.
 

Thus, oxidation is 5SO^{2-}_{3}+5H_{2}O \rightarrow 5SO^{2-}_{4}+10e^{-}+10H^{+}and

reduction is 2MnO_{4}^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O


 

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

 

2MnO_{4}^{-}+10e^{-}+16H^{+}+5SO_{3}^{2-} +5H_{2}O\rightarrow 2Mn^{2+}+8H_{2}O+5SO_{4}^{2-}+10e^{-}+10H^{+}

 

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
 

2MnO_{4}^{-}+6H^{+}+5SO_{3}^{2-} \rightarrow 2Mn^{2+}+3H_{2}O+5SO_{4}^{2-}

 

We will now verify whether all the charges are balanced.


So, LHS = 2\times-1+6+5\times-2= -6and the RHS = 2\times+2+3\times 0+5\times-2 = -6
Equal charges on both sides imply towards a balanced equation.


 

 (iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn_{2+} + Br_{2} + H_{2}O

 

As step (i), we will be generating the unbalanced skeleton i.e.,


 

(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O


As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.


 

Thus, oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-} and reduction is MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}


As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

As we can see that the atoms are balanced, therefore the reaction is going to be the same as the previous step.

 

Thus, oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-}  and reduction is  MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}


Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.


 

Thus, oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-}and

reduction is  MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}


 

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

 

Thus, the oxidation is 2Br^{-}\rightarrow Br_{2}+2e^{-} and

reduction is  MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}+4H_{2}O

 

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

 

Thus, oxidation is 10Br^{-}\rightarrow 5Br_{2}+10e^{-}and

reduction is  2MnO_{4}^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O


 

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
 

2MnO_{4}^{-}+5Br^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O+5Br_{2}+10e^{+}


Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

 

2MnO_{4}^{-}+10Br^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O+5Br_{2}

 

Thus, the charge on LHS and RHS is equal.

 

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