# Balance the following ionic equations $(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O$ $(iii) MnO^{-}_{4} + SO^{2-}_{3} + H^{+}\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O$ $(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O$

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

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 and Reduction is

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

Oxidation is   and Reduction is

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  and reduction is

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  and

reduction is

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  and

the reduction as

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.

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.

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.

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

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   and the reduction is

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

Thus, oxidation is   and the reduction is

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   and

reduction is

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 and

reduction is

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 and

reduction is

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.

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.

We will now verify whether all the charges are balanced.

So, the LHS = and the RHS =

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

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

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  and reduction is

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  and reduction is

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  and

reduction is

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 and

reduction is

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 and

reduction is

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.

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.

We will now verify whether all the charges are balanced.

So, LHS = and the RHS =
Equal charges on both sides imply towards a balanced equation.

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  and reduction is

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   and reduction is

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 and

reduction is

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  and

reduction is

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 and

reduction is

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.

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.

Thus, the charge on LHS and RHS is equal.

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