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Electrolysis As Oxidation-Reduction Reaction


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Strings (SiPjAjk) = S7P5A52     Base Sequence = 12735     String Sequence = 12735 - 5 - 52

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Electrolysis As Oxidation-Reduction Reaction

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Electrolysis As Oxidation-Reduction Reaction

Figure 14.56 is a simple conceptual illustration of the electrolytic decomposition of molten sodium chloride.
(a) Describe the half-reactions of the chemical reaction then derive arithmetically, the whole oxidation-reduction reaction from the half-reactions.

(b) Using Faraday's Law of electrolysis, calculate the amount of Chlorine (in grams), a chemist could produced from molten sodium chloride (NaCl) if she uses a current of 1 ampere for 5 minutes?

The strings:

S7P5A52 (Change - Chemical Change)

The math:
Pj Problem of Interest is of type change (chemical change).

Electrolysis As Oxidation Reduction Reaction

(a) Electrolysis is the use of electric current to cause a chemical reaction. The basic set-up (figure 14.56) is called an electrolytic cell. It consists of a voltage source connected to two electrodes immersed in a chemical substance. One of the electrodes, called the cathode, is negatively charged. The other, called the anode is positively charged. The substancce in which the electrodes are immersed is a solution or molten form of an electrolyte. An electrolyte is a substance whose aqueous solution conducts electricity. The conduction (called electrolytic conduction) that occurs in an electrolytic cell is due to the movement of positive ions (cations) and negative ions (anions) in the molten electrolyte. The cations are attracted to the cathode while the anions are attracted to the anode. In the electrolytic cell given in figure 14.56, the cations are sodium ions while the anions are chloride ions. So, the electrodes used must not react chemically with sodium or chloride ions.

The half-reactions are the reactions occuring at the electrodes:

Half-reaction at the cathode:
positively charged sodium ions are attracted to highly negatively charge cathode. Cathode electrons are at a higher potential energy than electrons in sodium atom. So electrons move from the cathode (high potential energy) to the sodium ions (lower potential energy). So at the cathode each sodium ion will be converted to a sodium atom by the addition of an electron and the half-reaction is:

Na+ + e- --------> Na -------(1)
This reaction at the cathode results in a gain of electron. So it is a reduction reaction.

Half-reaction at the anode:
Negatively charged chloride ions are attracted to highly positively charge anode. Anode electrons are at a lower potential energy than the electrons of the chloride ions. So, electrons move from the chloride ions to the anode. So at the anode, each chloride ion loses an electron and the half-reaction is:

2Cl- -------> Cl2 + 2e- -------(2)
This reaction at the anode results in a loss of electron. So, it is an oxidation reaction.

In general, oxidation always occurs at the anode and reduction always occurs at the cathode.

The half-reactions at the cathode and anode do not occur independently. So, they must be combine to get the whole oxidation-reduction reaction as follows:
reaction (1) is multiplied by 2 and added to reaction (2)

So, 2Na+ + 2e- + 2Cl- --------> 2Na + Cl2 + 2e-
So, 2Na+(l) + 2Cl-(l) --------> 2Na(l) + Cl2(g)

(b) Michael Faraday's (1791-1867 A.D) law of electrolysis relates the mass of substance produced at either eletrode during electrolysis to the amount of electricity used (expressed in Coulombs). The formula is as follows:

mass of substance = moles of electrons x formula mass of substance/moles of electrons transfered.
1 mole of electron is defined as the Farad and equals 96485 Coulombs of electricity.

The half-equation at the anode for the production of Cl2 is as follows:
2Cl- -------> Cl2 + 2e-
Quantity of electricity used in Coulombs = 1 ampere x 300secs = 300 Coulombs
So, moles of electrons = 300/96485
moles of electrons transferred = 2 (from the half-reaction)
So, mass of chlorine = (300/96485) x (70.9/2) = .00311 x 35.45 = 0.111g

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