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Oxidation-Reduction Reaction In A Voltaic Cell
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Oxidation-Reduction Reaction In A Voltaic Cell

Figure 14.63 is a simple illustration of a voltaic cell called the Daniell cell. Container 1 contains zinc sulphate solution in which a zinc electrode is immersed; container 2 contains copper(II) sulphate solution in which a copper electrode is immersed. A conducting wire connects the electrodes through a volmeter. The ends of an inverted U-shape salt bridge are immersed in the solutions respectively.

(a) Describe the half-reactions of the chemical reaction in the cell then derive arithmetically, the whole oxidation-reduction reaction from the half-reactions.

(b) Determine the approximate reading of the voltmeter.

The strings:

S₇P₅A₅₂ (Change - Chemical Change)

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

(a) A voltaic cell is a cell in which electric current is generated from a spontaneous chemical reaction. It is named after chemist Alessandro Volta (1745-1827) who in circa 1800 invented the first cell of this type. The voltaic cell is also called the galvanic cell in honor of Luigi Galvani (1737-1798), who is generally considered as the discoverer of electricity. The cell shown in figure 14.63 is called the Daniell cell. It is named after chemist Frederick Daniell who constructed the first version of the cell. The porous salt bridge prevents the mixing of the solution while providing a channel for the flow of sulphate ions between the solutions. As is seen in figure 14.63, the solutions are salts of their electrodes respectively. The flow of sulphate ions from container 2 allows for the resupply of depleted sulphate ions in container 1.

The half-reactions are the reactions occuring at the electrodes (the region in which a half reaction takes place is called a half-cell):

Half-reaction at the anode:
The zinc electrode is the anode. Zinc is a more active metal than copper because its valence electrons (outer electrons) have a higher potential energy than the valence electrons of copper. So, electrons flow from the zinc anode through the voltmeter to the copper cathode. So, at the anode, zinc looses electrons and the half-reaction is:

Zn -------> Zn²⁺ + 2e^- --------(1)
This reaction at the anode results in loss of electrons. So, it is an oxidation reaction.

Half-reaction at the cathode:
The copper electrode is the cathode. The potential energy of electrons of the copper electrode is higher than the potential energy of the electrons of the copper ions in the solution. So, the copper electrode transfers electrons to the copper ions and copper metal plates out on the cathode. So, at the cathode, copper gains electrons and the half-reaction is:

Cu²⁺ + 2e^- -------> Cu -------(2)
This reaction at the cathode results in a gain of electrons. So, it is a reduction reaction.

In general, oxidation always occurs at the anode and reduction always occurs at the cathode. The half reactions explain why there would be excess sulphate ions in the copper(II) sulphate solution and depleted sulphate ions in the zinc sulphate solution if there was no salt bridge to provide the channel for the flow of ions.

To get the whole (overall) oxidation-reduction reaction written as a net ionic equation:
Add half-reaction (1) to half-reaction (2)

So, Zn(c) + Cu²⁺(aq) + 2e^- --------> Zn²⁺(aq) + Cu(c) + 2e^-
So, Zn(c) + Cu²⁺(aq) --------> Zn²⁺(aq) + Cu(c)---------(3)

It is important to note that the oxidation-reduction reaction represented by (3) involves two electrodes. If only zinc is immersed in copper(II)sulphate solution, no electrical work will be done eventhough a redox reaction will still take place.

(b) The standard electrode potential is the difference in potential between the hydrogen half-cell and an electrode's half-cell. The standard electrode potential varies with temperature, pressure and concentration. There are tabulated values for 25⁰C, 1 atm and 1 Molarity.

To get voltmeter reading we add the standard electrode potentials of the half-cells.

Anode half-reaction:
Zn(c) -------> Zn²⁺⁺(aq) + 2e^-
Standard reduction potential from table = - 0.762 volts
Since this half-reaction is an oxidation reaction, we change -0.762 to +0.762.

Cathode half-reaction:
Cu²⁺(aq) + 2e^- -------> Cu(c)
Standard reduction potential from table = 0.340
Since this half-reaction is a reduction reaction, we do not need to change table value.

So, approximate voltmeter reading = 0.762 + 0.340 = 1.102 volts.

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