Problem 2: a.+3+1 -1 +1 -2 +1 +6 -2 +1 +3 +1 +3 -2 +1 +3 +1 -2
b. carbon in ethanol has an oxidation state of -1 but in acetic acid it is +3. Ethanol is oxidized CH3CH2OH(g) + H2O (l) → CH3CO2H(aq) + 4e- + 4 H+(aq) oxidation. To balance species in an acidic solution use the following: To the oxygen rich side add 2 H+ ions for each unbalanced oxygen and to the opposite add H2O for every oxygen deficient.
c. Chromium in dichromate has an oxidation state of +6 but changes to +3. Chromium is reduced. Balance the chromium atoms first, add the electrons, and finally add H+ ions to the oxygen rich side and water to the oxygen deficient side.
Cr2O72- (aq) + 14H+ (aq) + 6e- →. 2Cr3+ (aq) + 7H2O(l) reduction
d. Oxidizing agent: dichromate Cr2O72-(aq), Reducing agent: ethanol CH3CH2OH(g)
e. Electrons lost must equal electrons gained.
3 (CH3CH2OH(g) + H2O(l) → CH3CO2H(aq) + 4e- + 4 H+(aq))
3 CH3CH2OH(g + 3H2O(l) → 3CH3CO2H(aq + 12e- + 12 H+(aq)
2(Cr2O72- (aq) + 14H+(aq) + 6e- →. 2Cr 3+ (aq) + 7H2O(l))
2 Cr2O72- (aq) + 28H+(aq) + 12e- → 4Cr 3+ (aq) + 14H2O(l)
Add both half reactions and simplify by canceling or reducing like terms on both sides of the equation.
3 CH3CH2OH(g) + 3H2O(l) + 2 Cr2O72- (aq) + 28H+(aq) 16 H+ (aq)+ 12e- →
3 CH3CO2H(aq) + 12e- + 12 H+ + 4Cr 3+ (aq) + 14H2O(l) + 11H2O
balanced equation:
2Cr2O72- (aq) + 16H+ + (aq) + 3C 2H5OH(aq) →. 3CH3COOH(aq) + 4Cr 3+ (aq) + 11H2O(l)
Note: The equation above is completely balanced in terms of having an equal number of atoms as well as charges.
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