Question

3. a) Find the oxidation state of each of the labeled carbons and then label each reaction as "oxidation" or "reduction." (1 pt ea) OH OH

Answer 1

Answer: a) Left hand carbon is 0 oxidation state and right hand side carbon is -2 , so it is reduction reaction.

b) Left hand carbon is 0 oxidation state and right hand side carbon is -1 , so it is reduction reaction.

c) Left hand carbon is 0 oxidation state and right hand side carbon is +2 , so it is oxidation reaction.

d) Left hand carbon is -1 oxidation state and right hand side carbon is -2 , so it is reduction reaction.

EXPLANATION:

Oxidation number can be calculated as = Number of valence electrons in free atom - Number of valence electrons in bonded atom

For carbon atom number of valence electrons in free carbon atom is 4 .

If there is electronegativity difference between carbon atom and other bonded atom to carbon and if the other bonded atom is more electronegative than carbon then it will attract the shared pair of electrons towards itself more than bonded carbon and we have to not consider the shared pair of electrons betweewn carbon and bonded electronegative atom . And if carbon atom is attached to other carbon atom then we have to divide it by 2 as shared pair of electrons is shared equally beyween two bonded carbon atoms because of same electronegayivities . Lewis dot structure must be drawn.

If oxidation state of carbon atom increase from going left to right then it is oxidation and if decrease then it is reduction reaction

Full explanation is given below :

a).
a Loch Hese, cu . Since Ourgen is more electronegative than carbon so, it takes of up al all the two shared electrons towards itself. Elections between carbon and another carbon is Shened equally and we have to divide it by 2 and ore electronegative than Hydrogen, so carbon attracts . elections towards it. : So, for a. 6. is more Cartoon Orication state of corobon= 4-(3 =4-(1+1+2 + 2 + 2) = 4-u=6

of carbon is. onidation state OOCM3 H2 A So, for No Similarly for (in → A = / Oxi dation state of carbon = 4-( + ? 422) = 4-(1+1+2) = 4-16) = -2

For (i) on 2 . H Oncidation state of carbon =4- =4- (1+1+2) = 4-4 = 0 it it foo (6) N си, gcochy Oxidation state of carbon 4-(2 2 + 2 + 3 + 2) - 2 Unidation state carbon = 4- (1+1+1+2) = 4-5 = -1

@). Aco th ST Onicletion state of carbon = 4-{ 2 2 +2 +2} =4-(171+2) = 0

For (1) - Onication state of corobon = 4-6 2 2 + 2 } = 4-(1+1) = 2

(4). n => for cins olac Oniolation state of carbon = 4 (2+2 + =4-(1+1+1+2) = 4-5=-1 - For (6) - 3 Omnistohen state of Carden = 4-(3++2+2) Onidation state Carbon = + 2 +2 34-6) = -2
Summary

For,

a) Left hand carbon is 0 oxidation state and right hand side carbon is -2 , so it is reduction reaction.

b) Left hand carbon is 0 oxidation state and right hand side carbon is -1 , so it is reduction reaction.

c) Left hand carbon is 0 oxidation state and right hand side carbon is +2 , so it is oxidation reaction.

d) Left hand carbon is -1 oxidation state and right hand side carbon is -2 , so it is reduction reaction.