Question 2 Note: Do this one as you did Question 1. I'll link another Youtube video at the end of the problem statement. You should watch the video at the appropriate place, note down the voltage reading, & write it as your answer. To do this, we now need to switch the solutions in such a way that they will be as if the battery had run down. What I mean by this is that we are going to change the concentrations of the two solutions such that the product concentration will be greater than the reactant concentration. This is what happens when a battery runs down - the (reactant) will get smaller because it is being used up, and vice-versa for the product. Recall that Q. the reaction quotient, is defined as (Product]/ [Reactant], with each concentration raised to the appropriate power. For the zinc-copper cell, the Zn is the product and the Cu2 is the reactant. So Q will be [Zn2"]/ [Cu?). Adjust the concentrations of the two solutions as follows: For the copper solution, add 6 ml of the copper solution to 24 mL of deionized water. This will give you a Cu" concentration of 0.2 M. For the zinc solution, add 18 mL of the zinc solution to 12 ml of deionized water. This will give you a Zn"concentration of 0.6 M. Now, proceed exactly as you did in Part I, using the new solution concentrations. Since we are simulating a used battery here, your new voltage should be slightly lower than your voltage in Part I. A rough estimate is that it will be about 1.08 to 1.09 V. If your reading is off, again make sure the electrodes are clean, and then if the reading is still off you may wish to switch to a different multimeter. Here is the link to the Youtube video. Watch at 1:30 to 2:00 to see the reading they got (for this same experiment). https://www.youtube.com/watch?v-afEX2FD4Ado e My voltage reading for Part II is:
Basics in Organic Reactions Mechanisms
In organic chemistry, the mechanism of an organic reaction is defined as a complete step-by-step explanation of how a reaction of organic compounds happens. A completely detailed mechanism would relate the first structure of the reactants with the last structure of the products and would represent changes in structure and energy all through the reaction step.
Heterolytic Bond Breaking
Heterolytic bond breaking is also known as heterolysis or heterolytic fission or ionic fission. It is defined as breaking of a covalent bond between two different atoms in which one atom gains both of the shared pair of electrons. The atom that gains both electrons is more electronegative than the other atom in covalent bond. The energy needed for heterolytic fission is called as heterolytic bond dissociation energy.
Polar Aprotic Solvent
Solvents that are chemically polar in nature and are not capable of hydrogen bonding (implying that a hydrogen atom directly linked with an electronegative atom is not found) are referred to as polar aprotic solvents. Some commonly used polar aprotic solvents are acetone, DMF, acetonitrile, DMSO, etc.
Oxygen Nucleophiles
Oxygen being an electron rich species with a lone pair electron, can act as a good nucleophile. Typically, oxygen nucleophiles can be found in these compounds- water, hydroxides and alcohols.
Carbon Nucleophiles
We are aware that carbon belongs to group IV and hence does not possess any lone pair of electrons. Implying that neutral carbon is not a nucleophile then how is carbon going to be nucleophilic? The answer to this is that when a carbon atom is attached to a metal (can be seen in the case of organometallic compounds), the metal atom develops a partial positive charge and carbon develops a partial negative charge, hence making carbon nucleophilic.
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![Question 2
Note: Do this one as you did Question 1. I'II link another Youtube video at the end of the problem
statement. You should watch the video at the appropriate place, note down the voltage reading, &
write it as your answer.
To do this, we now need to switch the solutions in such a way that they will be as if the battery had
run down. What I mean by this is that we are going to change the concentrations of the two
solutions such that the product concentration will be greater than the reactant concentration. This is
what happens when a battery runs down the [reactant] will get smaller because it is being used up,
and vice-versa for the product. Recall that Q, the reaction quotient, is defined as [Product] /
[Reactant], with each concentration raised to the appropriate power. For the zinc-copper cell, the
Zn2+ is the product and the Cu2+ is the reactant. So Q will be [Zn2+]/[Cu2*].
Adjust the concentrations of the two solutions as follows:
For the copper solution, add 6 mL of the copper solution to 24 mL of deionized water. This will give
you a Cu2* concentration of 0.2 M.
For the zinc solution, add 18 mL of the zinc solution to 12 mL of deionized water. This will give you a
Zn2*concentration of 0.6 M.
Now, proceed exactly as you did in Part I, using the new solution concentrations. Since we are
simulating a used battery here, your new voltage should be slightly lower than your voltage in Part I.
A rough estimate is that it will be about 1.08 to 1.09 V. If your reading is off, again make sure the
electrodes are clean, and then if the reading is still off you may wish to switch to a different
multimeter.
Here is the link to the Youtube video. Watch at 1:30 to 2:00 to see the reading they got (for this
same experiment).
https://www.youtube.com/watch?v=afEX2FD4Ado e
My voltage reading for Part II is:
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