Question 1
Balance the following redox reaction by inserting the appropriate coefficients. HNO3 + H2S –> NO + S + H2O
Question 2
Question 3
Balance the following redox reaction by inserting the appropriate coefficients. [{MathJax fullWidth=’false’ \displaystyle \rm H_2O + Se + Cd^{2+} \to Cd + SeO_3^{2-} + H^+ }]
Balancing Chemical Reactions In balancing chemical reactions, both mass and charges must be balanced. The masses are balanced by adding numerical coefficients while charges are balanced by adding electrons on either the reactant or product side.
Question 4
(1)Why does the vanillin dissolve in an aqueous [{MathJax fullWidth=’false’ NaOH}] solution but not in water? (2)Is [{MathJax fullWidth=’false’ H^{-}}](hydride) used in excess in the vanillin reduction? (3)What is the gas that evolves during the work-up? (4)What is the purpose of adding aqueous [{MathJax fullWidth=’false’ HCl}] to the reaction mixture? (5)In the following synthesis, two reactions take place that you have carried out in the laboratory this semester. What are the two reactions and in what order do they take place? [{Image src=’currentproblem4122197253670502295.jpg’ alt=” caption=”}] [{Image src=’currentproblem2346377323461869060.jpg’ alt=” caption=”}] [{Image src=’currentproblem2752580628586330746.jpg’ alt=” caption=”}]
?!Vanillin Reduction:
In this reduction, vanillin is reduced to vanillyl alcohol by reaction with sodium borohydride in sodium hydroxide, followed by a reaction with hydrochloric acid.
Answer to question 1
steps to balance redox reactions
(1) identify the oxidation number of every atom
(2) determine which atoms are oxidized and which are reduced
(3) write half reactions and include electrons
(4) balance electrons in half reactions
(5) combine balanced half reactions and cancel electrons
(6) combine and or add counter ions
(7) add any remaining species THEN balance them
*** 1 ***
H in H2S is +1
S in H2S is -2
H in HNO3 is +1
N in HNO3 is +5
O in HNO3 is -2
S in “S” is 0
N in NO = +2
O in NO = -2
*** 2 ***
N went from +5 to +2 and was REDUCED.. (reduction = reduction in charge)
S went from -2 to 0 and was OXIDIZED.. (oxidation is the opposite of reduction.. increase in charge
no atoms of H nor O changed charges
*** 3 ***
1 S(-2) —> 1 S(0) + 2e’s
1 N(+5) + 3 e’s —> 1 N(+2)
*** 4 ***
3x the first + 2x the second balances the e’s
3 S(-2) —> 3 S(0) + 6e’s
2 N(+5) + 6 e’s —> 2 N(+2)
*** 5 ***
3 S(-2) + 2 N(+5) + 6 e’s —> 3 S(0) + 6e’s + 2 N(+2)
3 S(-2) + 2 N(+5) —> 3 S(0) + 2 N(+2)
*** 6 ***
3 H2S + 2 HNO3 —> 3 S + 2 NO
*** 7 ***
notice you still have 3×2 + 2×1 = 8 H’s on the left and NONE on the right?
notice you have 6 O’s on the left and only 2 on the right?
what do you suppose 8H’s and 6-2 = 4 O’s make? that’s right.. WATER so we need to add water on the right
3 H2S + 2 HNO3 —> 3 S + 2 NO + __ H2O
and finally balancing the H’s
3 H2S + 2 HNO3 —> 3 S + 2 NO + 4 H2O
Answer to question 2
Fe in Fe+3 has oxidation state of +3
So, Fe in Fe+3 is reduced to Fe+2
N in NO2- has oxidation state of +3
N in NO3- has oxidation state of +5
So, N in NO2- is oxidised to NO3-
Reduction half cell:
Fe+3 + 1e- –> Fe+2
Oxidation half cell:
NO2- –> NO3- + 2e-
Balance number of electrons to be same in both half reactions
Reduction half cell:
2 Fe+3 + 2e- –> 2 Fe+2
Oxidation half cell:
NO2- –> NO3- + 2e-
Lets combine both the reactions.
2 Fe+3 + NO2- –> 2 Fe+2 + NO3-
Balance Oxygen by adding water
2 Fe+3 + NO2- + H2O –> 2 Fe+2 + NO3-
Balance Hydrogen by adding H+
2 Fe+3 + NO2- + H2O –> 2 Fe+2 + NO3- + 2 H+
This is balanced chemical equation in acidic medium
Answer:
2 Fe3+ + NO2– + H2O —> 2 Fe2+ + 2 H+ + NO3–
Answer to question 3
Answer to question 4
a) Reaction of vanillin with aqueous sodium hydroxide is involved in the process of reduction of vanillin by sodium borohydride. The alcohol group on phenol of vanillin has an acidic proton. Sodium hydroxide being the conjugate base of water reacts with vanillin and deprotonates the alcoholic group attached to phenol. This salt formed is soluble in water. Deprotonation of vanillin is done so that on the addition of sodium borohydride there is an attack on the carbonyl group instead of hydrogen of the alcoholic group, resulting in the formation of hydrogen gas.
b) Gain in electrons or hydrogen causes reduction reaction. There are four hydrogen atoms present in sodium borohydride (reducing agent). Since hydride ions (nucleophile) have filled 1s orbital and carbon has comparatively diffused 2p orbital (carbon-oxygen double bond), hydrid orbital is too small for interaction with orbital of carbon. Excess of hydride is used in vanillin reduction. Hence, excess of hydride would increase the reaction rate.
