Question

For the reacton in an acidic medium :
$C_2{H}_{5}OH+C{r}_2{O}_7^{2-}\to C{H}_{3}COOH+C{r}^{3+}$
Calculate the stoichiometric ratio of $C_2{H}_{5}OH$ to $C{r}_2{O}_7^{2-}$ when the reaction is balanced by oxidation number method

• A. 1
• B. 1.5
• C. 2
• D. 0.5

Answer 1

The correct option is B 1.5

$C_2{H}_{5}OH+C{r}_2{O}_7^{2-}\to C{H}_{3}COOH+C{r}^{3+}$

Oxidation state of $C$ in $C_2{H}_{5}OH=-2$
Oxidation state of $C$ in $C{H}_{3}COOH=0$
Oxidation state of $Cr$ in $C{r}_2{O}_7^{2-}=+6$
Oxidation state of $Cr$ in $C{r}^{3+}=+3$

$C_2{H}_{5}OH$ is undergoing oxidation and is the reducing agent
$C{r}_2{O}_7^{2-}$ is undergoing reduction and is oxidising agent.
For n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$

$n_f$ of $C_2{H}_{5}OH=4$
$n_f$ of $C{r}_2{O}_7^{2-}=6$

Ratio of oxidation states is $2:3$

Cross multiplying these with $n_f$ of each other.
we get,
$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}\to C{H}_{3}COOH+C{r}^{3+}$

Balancing the elements other than oxygen and hydrogen on both sides,

$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}\to 3C{H}_{3}COOH+4C{r}^{3+}$

Adding the $H_{2}O$ to balance the oxygen,


$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}\to 3C{H}_{3}COOH+4C{r}^{3+}+11H_{2}O$

Adding $H^{+}$ to balance hydrogen,

$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}+16H^{+}\to 3C{H}_{3}COOH+4C{r}^{3+}+11H_{2}O$

Balance charge
total charge in reactant side=+12
total charge in product side=+12

$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}+16H^{+}\to 3C{H}_{3}COOH+4C{r}^{3+}+11H_{2}O$)

This is the final balanced equation.
So, required ratio is $\frac{n_{C_2{H}_{5}OH}}{n_{C{r}_2{O}_7^{2-}}}=\frac{3}{2}=1.5$