Question

One method for measuring the alcohol content of a liquid is to titrate the ethanol (C2H5OH) with chromate (CrO4^2-). These reagents react to become Co2 and Cr^+3 respectively. Write the balanced equation for this redoc process under acidic conditions. If a 1.86g sample of an alcoholic beverage is titrated with 25.93 mL of 0.250 M sodium chromate, what is the percent by mass of alcohol in this beverage?

Answer 1

The titration reaction of C2H5OH with CrO4^2- can be written as

To balance the reaction, we will first separate them into half reactions:

Oxidation:

Reduction:

Now, lets balance the oxidation half first

We see that there are 2 C on left and 1 C on right. Hence, we multiply 2 on CO2 to balance the number of C on both side.

Now, we have 4 O on right side and 1 O on left side. Hence, we add 3 H2O on left to balance the number of O atoms.

Now, we have 12 H on left and no H on right. Hence, we add 12 H+ ions on right side to balance the number of H atoms.

Now, the left side has no charge and right side has 12 + charge. Hence, we add 12 electrons on the right to balance the charges.

Hence, the balanced oxidation half is

Now, we will balance the reduction half

Left side has 4 O and right side has no O atoms. Hence, we add 4H2O on the right side to balance the number of O atoms.

Now, we have 8 H on right and none on left. Hence, we add 8 H+ on left to balance the number of H atoms.

Now, we have -2+8 =+6 charge on left and +3 charge on right. Hence, we add 3 electrons on the left to balance the charges.

Hence, the balanced reduction half reaction is

Now, we can multiply 4 throughout in reduction half to make the number of electrons same in both the half reactions.

Now, we can add the half reactions and cancel the exact species from both side to get the balanced redox reaction.

+

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Now, we can cancel 3H2O and 12 H+ from both sides to get the final balanced redox reaction

Note that 4 moles of chromate ion reacts with 1 mol of C2H5OH.

Now, we have used 25.93 mL of 0.250 M sodium chromate .

Hence, the amount of chromate ions in the solution is

We know that 4 moles of chromate ions react with 1 mol of C2H5OH.

Hence, number of moles of C2H5OH that must be in the beverage is

Now, molar mass of C2H5OH = 46.07 g/mol

Hence, mass of C2H5OH present in the beverage is

The mass of the beverage sample = 1.86 g

hence, the percent by mass of the alcohol C2H5OH in the beverage can be calculated as

Hence, the percent by mass of the alcohol in the beverage is about 4.01%.