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Ch.20 - Electrochemistry

Chapter 20, Problem 106c

Disulfides are compounds that have S ¬ S bonds, like peroxides have O ¬ O bonds. Thiols are organic compounds that have the general formula R ¬ SH, where R is a generic hydrocarbon. The SH- ion is the sulfur counterpart of hydroxide, OH-. Two thiols can react to make a disulfide, R ¬ S ¬ S ¬ R. (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols?

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Hey everyone today, we're looking at the reaction between oxygen and to hide to form parasitic acid. And we're being asked if oxygen is being reduced or oxidized in this reaction. So before anything to find whether in a molecule element is being oxidized or reduced, we need to find out what is its oxidation number before and after the reaction has taken place. So let's first look at before the reaction has taken place. We have oxygen and we have a seat allowed to hide CH three C. H. O. So in the case of oxygen, usually oxygen, according to our oxidation rules would have a oxidation state of negative two. However, oxygen is in its most standard state here. If it is in a sterile environment, oxygen would form 02 naturally. So the oxidation state of this of 02 would be zero And for Ch. three. Oh, The oxidation state here is also zero. And we don't have to really worry about it because the only oxidation state that were truly worried about is for the oxygen's that are reacting with the city to hide the oxidation state of the O. In a city called a hide is negative two. But that's not what we're worried about. So with that in mind we can go ahead and start writing out the proceeding acid. Now in this case it would actually be a little easier to draw the whole molecule. So let me do that real quick. We have a carbon bonded to three hydrogen. The CHC on the left bonded to A C. Excuse me, it'll be double bonded to an O. And bonded to 00. And H. So the key thing to notice here is that specifically these two oxygen's are the two oxygen's that are from here, all the rest belong to the original aside to hide. So taking a look here in order to identify what is the oxidation state of each of these oxygen's here, we first need to identify the oxidation states of every other molecule here. Now, according to our scalability rules when hydrogen is bonded to a nonmetal, It will have a oxidation state of plus one. So that will go for every single oxygen in this molecule. Right. This a little neater plus one and plus one oxygen will have a oxidation state of negative two when bonded to a non metal, as we can see here. And the carbons will now vary depending on what they are attached to. Now keep in mind that parasitic acid is a neutral molecule, which means it has a total oxidation state of zero, which means all the oxidation states of each molecule needs to add up to zero. So let's break this down into two halves. We have this left half this metal group and then we have this right half. So taking a look at the left half. First, we have plus three as a total oxidation state so far due to the three hydrogen, which means in order for this to equal zero, the carbon must have a total charge of negative three. Pretty simple because negative three plus three is equal to zero. However, for the right half it's a little more complicated now, we have negative two here and we have a plus one. So like so far, the total charge on the right side is negative one due to the negative two and the plus one from the hydrogen. However, usually oxygen have an oxidation state um negative two here, which means if these were to have a negative two oxidation state, that would be negative two and negative two. And we would end up with negative five as a total here, What do you call -5. However, we are dealing with a neutral molecule and carbon can only have a total oxidation state Of -4 or Plus four. That is its range. Let me write this a little clear over here. This carbon has a negative three but this carbon can only have a maximum of negative four as an oxidation state, Which means in order for this to be a neutral molecule or sorry negative for this needs to be plus four plus four. However, we need to ensure that the right side can not have a charge a charge at all. It needs to still be zero. So to do this an easy way would be to put a charge of plus three on the carbon here as well. What this does is it brings us to a total charge of positive two on the right side, and oxygen reacting with. It means that each oxygen therefore has a charge of negative one, which will bring us to A total charge of zero. So the oxidation state of zero or sorry, of oxygen goes from zero two negative one. This negative one. So since it is gaining electrons here, oxygen is being reduced because, remember oil rig oxidation is loss of electrons, reduction is gain of electrons. So since it's going from zero to negative one, oxygen is being reduced reduced. I hope this helps. And I look forward to seeing you all in the next one.
Related Practice
Textbook Question
(b) Given the following reduction potentials, calculate the standard emf of the cell: Cd1OH221s2 + 2 e- ¡ Cd1s2 + 2 OH-1aq2 E°red = -0.76 V NiO1OH21s2 + H2O1l2 + e- ¡ Ni1OH221s2 + OH-1aq2 E°red = +0.49 V
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Textbook Question

The capacity of batteries such as the typical AA alkaline battery is expressed in units of milliamp-hours (mAh). An AA alkaline battery yields a nominal capacity of 2850 mAh. (b) The starting voltage of a fresh alkaline battery is 1.55 V. The voltage decreases during discharge and is 0.80 V when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

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Textbook Question

Disulfides are compounds that have S ¬ S bonds, like peroxides have O ¬ O bonds. Thiols are organic compounds that have the general formula R ¬ SH, where R is a generic hydrocarbon. The SH- ion is the sulfur counterpart of hydroxide, OH-. Two thiols can react to make a disulfide, R ¬ S ¬ S ¬ R. (b) What is the oxidation state of sulfur in a disulfide?

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Textbook Question

Calculate the number of kilowatt-hours of electricity required to produce 1.0 * 103 kg (1 metric ton) of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V and the process is 45% efficient.

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Textbook Question

The Haber process is the principal industrial route for converting nitrogen into ammonia: N2(g) + 3 H2(g) → 2 NH3(g) (b) Using the thermodynamic data in Appendix C, calculate the equilibrium constant for the process at room temperature.

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Textbook Question

Aqueous solutions of ammonia 1NH32 and bleach (active ingredient NaOCl) are sold as cleaning fluids, but bottles of both of them warn: 'Never mix ammonia and bleach, as toxic gases may be produced.' One of the toxic gases that can be produced is chloroamine, NH2Cl. (b) What is the oxidation number of chlorine in chloramine?

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