Question

What are the stoichiometric coefficients in the balanced redox reaction only for the species in the unbalanced reaction respectively.This reaction is used by the Breathalyzer to detect ethyl alcohol in a driver's breath?
$C_2{H}_{5}OH+C{r}_2{O}_7^{2-}\to C{H}_{3}COOH+C{r}^{3+}$

• A. Reactants: 3, 4; products: 3, 8
• B. Reactants: 2, 2; products: 2, 4
• C. Reactants: 3, 1; products: 3, 2
• D. Reactants: 3, 2; products: 3, 4

Answer 1

The correct option is D Reactants: 3, 2; products: 3, 4

Steps for Balancing redox reactions:

• Identify the oxidation and reduction halves.
• Find the oxidising and reducing agent.
• Find the n-factor of oxidising and reducing agents.
• Cross multiply the oxidising and reducing agent with the simplified n-factor of each other.
• Balance the atoms other than oxygen and hydrogen.
• Balance oxygen atoms.
• Balance hydrogen atoms.
• Balance charge

formula used for the n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$

Taking the given equation and following the above mentioned steps:

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

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 $C{r}^{3+}=+3$

Cleary, $C_2{H}_{5}OH$ is undergoing oxidation and $C{r}_2{O}_7^{2-}$ is undergoing reduction.
using the formula of n-factor given above,

$n_f$ 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,

LHS of the equation contains 17 oxygens while RHS contains only 6 oxygens, so adding 11 $H_{2}O$ to RHS.
$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}\to 3C{H}_{3}COOH+4C{r}^{3+}+11H_{2}O$
adding $H^{+}$ to balance hydrogen,

The LHS of the equation contains 18 hydrogens while RHS contains 34 hydrogens, so adding 16$H^{+}$ to LHS.
$3C_2{H}_{5}OH+2C{r}_2{O}_7^{2-}+16H^{+}\to 3C{H}_{3}COOH+4C{r}^{3+}+11H_{2}O$

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 coefficients, reactants: 3, 2; products: 3, 4