Question

Balance the following equation by oxidation number method.
(i) ${K_2}C{r_2}{O_7} + KI + {H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {I_2} + {H_2}O$
(ii) $KMn{O_4} + N{a_2}S{O_3} \to Mn{O_2} + N{a_2}S{O_4} + KOH$
(iii) $Cu + HN{O_3} \to Cu{(N{O_3})_2} + N{O_2} + {H_2}O$
(iv) $KMn{O_4} + {H_2}{C_2}{O_4} + {H_2}S{O_4} \to {K_2}S{O_4} + MnS{O_4} + C{O_2} + {H_2}O$

Answer 1

Hint In order to balance the given chemical equation by oxidation number method, we must first have an idea about what an oxidation number method is. In the oxidation number method, we will assign the oxidation number to all the reactants and products present and then finally balance the equation.

Complete step by step solution:
Let us move onto the given problem.
(i) ${K_2}C{r_2}{O_7} + KI + {H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {I_2} + {H_2}O$
Let us first write the unbalanced equation.
${K_2}C{r_2}{O_7} + KI + {H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {I_2} + {H_2}O$
Now we find the oxidation number of the elements.
The oxidation number of I in $KI$ is -1.
The oxidation number of I in ${I_2}$ is 0.
As the oxidation number is increased oxidation reaction is taking place.
The oxidation number of Cr in ${K_2}C{r_2}{O_7}$ is +6.
The oxidation number of Cr in $C{r_2}{(S{O_4})_3}$ is +3.
As the oxidation number is decreased, a reduction reaction is taking place.
First is we have to balance iodine, so multiply $KI$ by 6 and ${I_2}$ by 3. Therefore, the iodine on both sides will become equal.
${K_2}C{r_2}{O_7} + 6KI + {H_2}S{O_4} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3{I_2} + {H_2}O$
Now we have to balance the potassium on both sides. For that multiply ${K_2}S{O_4}$ by 4.
${K_2}C{r_2}{O_7} + 6KI + {H_2}S{O_4} \to 4{K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3{I_2} + {H_2}O$
Now balance the $SO_4^{2 - }$ ion on both sides. Since the right side has 7 $SO_4^{2 - }$ ions, we have to multiply ${H_2}S{O_4}$ by 7.
${K_2}C{r_2}{O_7} + 6KI + 7{H_2}S{O_4} \to 4{K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3{I_2} + {H_2}O$
Finally balance the hydrogen and oxygen atoms on both sides. For that multiply ${H_2}O$ by 7. The balanced chemical equation is given as:
${K_2}C{r_2}{O_7} + 6KI + 7{H_2}S{O_4} \to 4{K_2}S{O_4} + C{r_2}{(S{O_4})_3} + 3{I_2} + 7{H_2}O$

(ii) $KMn{O_4} + N{a_2}S{O_3} \to Mn{O_2} + N{a_2}S{O_4} + KOH$
First write the unbalanced equation.
$KMn{O_4} + N{a_2}S{O_3} \to Mn{O_2} + N{a_2}S{O_4} + KOH$
Now we find the oxidation number of the elements.
The oxidation number of Mn in $KMn{O_4}$ is 7.
The oxidation number of Mn in $Mn{O_2}$ is 4.
As the oxidation number is decreased, a reduction reaction is taking place.
The oxidation number of S in $N{a_2}S{O_3}$ is 4.
The oxidation number of S in $N{a_2}S{O_4}$ is 6.
As the oxidation number is increased oxidation reaction is taking place.
Now we have to make the number of Mn on both sides equal so multiply $KMn{O_4}$ and $Mn{O_2}$ by 2 and then multiply $N{a_2}S{O_3}$ and $N{a_2}S{O_4}$ by 3.
$2KMn{O_4} + 3N{a_2}S{O_3} \to 2Mn{O_2} + 3N{a_2}S{O_4} + KOH$
Finally, we have to make the number of potassium equal to multiply$KOH$ by 2. Then to make the number of oxygen and hydrogen equal we have to add a water molecule on the left side.
$2KMn{O_4} + 3N{a_2}S{O_3} + {H_2}O \to 2Mn{O_2} + 3N{a_2}S{O_4} + 2KOH$
The above given equation is the balanced equation

(iii) $Cu + HN{O_3} \to Cu{(N{O_3})_2} + N{O_2} + {H_2}O$
First write the unbalanced equation.
$Cu + HN{O_3} \to Cu{(N{O_3})_2} + N{O_2} + {H_2}O$
Now we find the oxidation number of the elements.
The oxidation number of Cu in $Cu$ is 0
The oxidation number of Cu in $Cu{(N{O_3})_2}$ is 2.
As the oxidation number is increased oxidation reaction is taking place.
The oxidation number of N in $HN{O_3}$ is 5.
The oxidation number of N in $N{O_2}$ is 4.
As the oxidation number is decreased, a reduction reaction is taking place.
First, we have to make the nitrogen atoms equal on both sides, so multiply $HN{O_3}$ by 4 and $N{O_2}$ by 2.
$Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + {H_2}O$
Finally, to balance the hydrogen and oxygen atom, we have to multiply${H_2}O$ by 2.
$Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O$
The above given equation is a balanced equation.

(iv) $KMn{O_4} + {H_2}{C_2}{O_4} + {H_2}S{O_4} \to {K_2}S{O_4} + MnS{O_4} + C{O_2} + {H_2}O$
First write the unbalanced equation.
$KMn{O_4} + {H_2}{C_2}{O_4} + {H_2}S{O_4} \to {K_2}S{O_4} + MnS{O_4} + C{O_2} + {H_2}O$
Now we find the oxidation number of the elements.
The oxidation number of Mn in $KMn{O_4}$ is 7
The oxidation number of Mn in $MnS{O_4}$ is 2
As the oxidation number is decreased, a reduction reaction is taking place.
The oxidation number of C in ${H_2}{C_2}{O_4}$ is 3.
The oxidation number of C in $C{O_2}$ is 4.
As the oxidation number is increased oxidation reaction is taking place.
First, we have to balance the C on both sides, so we have to multiply ${H_2}{C_2}{O_4}$ by 5 and the multiply $C{O_2}$ by 10.
$KMn{O_4} + 5{H_2}{C_2}{O_4} + {H_2}S{O_4} \to {K_2}S{O_4} + MnS{O_4} + 10C{O_2} + {H_2}O$
Now we have to balance K on both sides for that we have to multiply $KMn{O_4}$ by 2. Then balance Mn on both sides, we have to multiply $MnS{O_4}$ by 2.
$2KMn{O_4} + 5{H_2}{C_2}{O_4} + {H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 10C{O_2} + {H_2}O$
Then to balance $SO_4^{2 - }$ on both sides, we have to multiply ${H_2}S{O_4}$ by 3.
$2KMn{O_4} + 5{H_2}{C_2}{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 10C{O_2} + {H_2}O$
Then finally to balance hydrogen and oxygen, we have to multiply ${H_2}O$ by 8.
$2KMn{O_4} + 5{H_2}{C_2}{O_4} + 3{H_2}S{O_4} \to {K_2}S{O_4} + 2MnS{O_4} + 10C{O_2} + 8{H_2}O$
The above given equation is a balanced equation.

Note: We have to remember that other than the oxidation number method, there is also one more method to balance the chemical equation. This method is known as the ion-electron method. In this method, we will take the half reaction separately and then balance it.