Question

During reflux, you oxidized ethanol to acetic acid, using postassium dichromate.

A) How many grams of ethanol can be oxidized by 100.00mL of a 0.750M K₂Cr₂O₇ solution?

B) In a similar reaction, a chemist oxidizes methanol (CH₃OH) to formic acid (HCOOH), also using K₂Cr₂O₇. Starting with half reactions, write a balanced redox equation for the reaction performed by the chemist. Show all steps

i. Which reactant is the oxidizing agent

ii. Which reactant is the reducing agent?

Answer 1

A) The chemical reaction for the oxidation under acidic conditions are,

3C2H5OH + 2K2Cr2O7 + 8H2SO4 ---> 3CH3CO2H + 2Cr2(SO4)3 + 2K2SO4 + 11H2O

So, 3 moles of ethanol gets oxidized by 2 moles of K2Cr2O7 to form 3 moles of acetic acid

moles of K2Cr2O7 = molarity x volume = 0.750 M x 0.1 L = 0.075 mols

So, moles of ethanol = 0.075 x 3/2 = 0.1125 mols

grams of ethanol getting oxidized = 0.1125 x 46.06844 = 5.183 g

B) Here CH3OH is getting oxidized and K2Cr2O7 is getting reduced

C in CH3OH is -2 goes to 0 in HCOOH by loosing 2e-

and, Cr in K2Cr2O7 is +6 goes to +3 by gaining 3e-

oxidation half cell,

CH3OH ---> HCOOH

balance O by adding H2O on the other side,

CH3OH + H2O ---> HCOOH

balance H by adding H+ on the other side,

CH3OH + H2O ---> HCOOH + 4H+

Add electrons,

CH3OH + H2O ---> HCOOH + 4H+ + 2e-

Reduction half cell,

Cr2O7^2- ----> 2Cr^3+

balance Cr,

2Cr2O7^2- ----> 2Cr^3+

balance O by adding H2O,

2Cr2O7^2- ----> 2Cr^3+ + 14H2O

balance e-'s

2Cr2O7^2- + 6e- ----> 2Cr^3+ + 14H2O

Add both equations, multiply the first equation with 3,

3CH3OH + 3H2O + 2Cr2O7^2- + 6e- ---> 3HCOOH + 12H+ + 6e- + 2Cr^3+ + 14H2O

cancel the common terms,

3CH3OH + 2Cr2O7^2- -----> 3HCOOH + 2Cr^3+ + 11H2O

Balance the other atoms (in acidic medium),

3CH3OH + 2K2Cr2O7 + H2SO4 -----> 3HCOOH + 2Cr2(SO4)3 + 11H2O + K2SO4

balance K and SO4 on both sides,

3CH3OH + 2K2Cr2O7 + 8H2SO4-----> 3HCOOH + 2Cr2(SO4)3 + 11H2O + 2K2SO4

Wil be the final balanced equation

i) K2Cr2O7 is the oxidizing agent.

ii) CH3OH is the reducing agent.