The blond alcohol (C 2 H 5 OH) level can be determined by titrating a sample of blond plasma with an acidic potassium dichromate solution, resulting in the production of Cr 3+ ( aq ) and carbon dioxide. The reaction can be monitored because the dichromate ion (Cr 2 O 7 2− ) is orange in solution. and the Cr 3+ ion is green. 'The unbalanced redox equation is Cr 2 O 7 2 + ( a q ) + C 2 H 5 OH ( a q ) → Cr 3 + ( a q ) + CO 2 ( g ) If 31.05 mL of 0.0600 M potassium dichromate solution is required to titrate 30.0 g blood plasma. determine the mass percent of alcohol in the blood.

Interpretation:

The determination of the blood plasma level by the titration of a sample of blood with an acidic Potassium dichromate solution resulting in the production of green colored Cr3+ is given. The mass percent of alcohol present in the given mass of blood plasma is to be calculated.

Concept introduction:

The mass percent is used to represent the contribution of an element in a given compound or a component in a given mixture. The mass percent is determined as,

Mass percent=Grams of soluteGrams of solution×100

To determine: The mass percent of alcohol present in the given mass of blood plasma by titrating against given volume of Dichromate of certain concentration.

Given,

The volume of Potassium dichromate is 31.05 ml .

The concentration of Potassium dichromate is 0.0600 M .

The mass of blood plasma is 30.0 g .

The given reaction is,

Cr2O72−(aq)+C2H5OH(aq)→Cr3+(aq)+CO2(g)

The reduction half-reaction is,

Cr2O72−(aq)→Cr3+(aq)

The oxidation of Chromium changes from +6 to +3 .

The oxidation half-reaction is,

C2H5OH(aq)→CO2(g)

The oxidation state of Carbon changes from −2 to +4 .

Balance all the elements except Hydrogen and oxygen in reduction half-reaction,

Cr2O72−(aq)→2Cr3+(aq)

Balance the Oxygen atoms by adding water molecules to the right hand side of the equation,

Cr2O72−(aq)→2Cr3+(aq)+7H2O(l)

Balance the number of Hydrogen atoms by adding H+ to the left hand side of the equation,

Cr2O72−(aq)+14H+(aq)→2Cr3+(aq)+7H2O(l)

Balance the charge by adding the appropriate number of electrons,

Cr2O72−(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(l)

Balance all the elements except Hydrogen and oxygen in oxidation half-reaction,

C2H5OH(aq)→2CO2(g)

Balance the Oxygen atoms by adding water molecules to the left hand side of the equation,

C2H5OH(aq)+3H2O(l)→2CO2(g)

Balance the number of Hydrogen atoms by adding H+ to the right hand side of the equation,

C2H5OH(aq)+3H2O(l)→2CO2(g)+12H+(aq)

Balance the charge by adding the appropriate number of electrons,

C2H5OH(aq)+3H2O(l)→2CO2(g)+12H+(aq)+12e−

Multiply the reduction half-reaction with a coefficient of 2 and then add both the oxidation and reduction half-reaction,

2Cr2O72−(