Table of contents |
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Half Cell Reactions |
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Solved Examples |
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The Hydrogen Half Cell |
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Standard Reduction Potential |
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Example 1: Write the anode and cathode reactions for a galvanic cell that utilizes the reaction
Ni(s) + 2Fe3+ → Ni2+ + 2Fe2+
Ans: Oxidation takes place at the anode, and the electrode must be Ni|Ni2+,
and the reduction occurs at the cathode: Fe3+, Fe2+,
2Fe3+ + 2e− → 2Fe2+
For every Ni atom oxidized, two Fe3+ ions are reduced. The electrons from the Ni metal will flow from the anode, pass the load, and then carry out the reduction at the surface of the cathode to reduce the ferric (Fe3+) ions to ferrous ions. In the meantime the ions in the solution move accordingly to keep the charges balanced.
Discussion
The galvanic cell is:
where represents a solution containing two types of ions. An inert Pt electrode is placed in the solution to provide electrons for the reduction.
Example 2: The charge on an electron is 1.602 x 10-19 C (coulomb). What is the charge on 1 mole of electrons?
Ans: The charge on one mole (Avogadro's number of) electrons is called a Faraday (F).
F = (6.022045 × 1023/mol) × (1.602 × 10−19C)
= 96485 C/mol (1), (2)
The chemical history involving the determination of Avogadro's number, and the charge on an electron, and how the two values agree with each other is very interesting.
Example 3: A galvanic cell with a voltage of 1.1 V utilizes the reaction
Zn + Cu2+ → Cu + Zn2+
as a source of energy. If 6.3 g of Cu and 11 g Zn are used, what is the maximum usable energy in this battery?
Ans: The 6.3 g Cu and 11 g Zn correspond to 0.10 and 0.17 mol of Cu and Zn respectively. Thus, Cu is the limiting reagent, and 0.10 mol corresponds to a charge of 2×96485×0.10 C (2 significant figures). The maximum available energy is then
Max. Energy = (1.1V) (96485C) (2) (0.10)
= 22000J (1J = 1VC) (3),(4)
Discussion
This energy corresponds to 2500 cal, which is enough to bring 25 g water from 273 K to its boiling point (373 K). Another way of looking at it: 22000 J is enough energy to send a 20-gram rocket to a height of 56 m.
Example 4: If the galvanic cell of Example 3 is used to power a calculator, which consumes 1 mW, how long theoretically will the battery last in continuous operation?
Ans: Power consumption of 1 mW is equivalent to 0.001 J/sec.
(5), (6), (7)
This is a realistic example. Most recent calculators use very little power. I noted that a SHARP programmable calculator uses 15 mW, a Casio calculator uses 0.5 mW, and an HP 25 uses 500 mW.
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Half-Cell Reaction
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[Question: 952230]
1. What is a half cell reaction in chemistry? | ![]() |
2. What is the significance of standard reduction potential in half cell reactions? | ![]() |
3. How is the hydrogen half cell reaction represented? | ![]() |
4. What is the role of a half cell in an electrochemical cell? | ![]() |
5. How can half cell reactions be used to determine standard reduction potentials? | ![]() |