Step 1: Understanding the Concept:
This problem applies Faraday's laws of electrolysis.
Electrochemical reactions involve the transfer of electrons. Oxidation occurs at the anode, where species lose electrons.
The quantity of electrical charge (\(Q\)) required for a redox reaction is directly proportional to the amount of substance reacting and the number of moles of electrons transferred per mole of substance.
The charge of one mole of electrons is defined as Faraday's constant (\(F\)), which is approximately 96,500 Coulombs.
Step 2: Key Formula or Approach:
The equation relating charge and electron transfer is:
\[ Q = n \cdot F \]
Where:
\(Q\) is the total charge in Coulombs (C).
\(n\) is the number of moles of electrons transferred in the balanced half-reaction.
\(F\) is Faraday’s constant ($96,500 C/mol$).
Step 3: Detailed Explanation:
1. Write the balanced oxidation half-reaction for water:
Water is oxidized to oxygen gas at the anode. The balanced equation is:
\[ 2H_{2}O \rightarrow O_{2} + 4H^{+} + 4e^{-} \]
2. Determine the stoichiometry:
According to the balanced equation, 2 moles of \(H_{2}O\) are oxidized to produce 1 mole of \(O_{2}\), releasing 4 moles of electrons (\(e^{-}\)).
The question asks for the charge required to oxidize 1 mole of \(H_{2}O\).
If 2 moles of \(H_{2}O\) require 4 moles of electrons, then 1 mole of \(H_{2}O\) requires:
\[ n = \frac{4}{2} = 2 \text{ moles of electrons.} \]
3. Calculate the charge:
Now, substitute \(n = 2\) and \(F = 96500\) into the formula:
\[ Q = 2 \times 96500 \]
\[ Q = 193000 \text{ Coulombs.} \]
4. Convert to scientific notation:
\[ Q = 1.93 \times 10^{5} \text{ C.} \]
This calculation shows that a substantial amount of electricity is needed to electrolyze water and release oxygen.
Step 4: Final Answer:
The quantity of electricity required is \(1.93 \times 10^{5}\) C.