Problem 98
Question
In electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what is liberated at anode? (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{SO}_{4}^{2-}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{O}_{2}\)
Step-by-Step Solution
Verified Answer
Oxygen \( \mathrm{O}_2 \) is liberated at the anode.
1Step 1: Identify Components in Solution
In dilute \( \mathrm{H}_2 \mathrm{SO}_4 \), the solution dissociates into \( \mathrm{H}^+ \) and \( \mathrm{SO}_4^{2-} \) ions.
2Step 2: Determine Anode Reaction
At the anode (positive electrode), oxidation occurs. The potential reactions could involve the oxidation of water molecules or \( \mathrm{SO}_4^{2-} \) ions. However, \( \mathrm{SO}_4^{2-} \) ions do not get oxidized easily in this context.
3Step 3: Consider Reaction of Water
Since \( \mathrm{SO}_4^{2-} \) ions are stable and will not oxidize, water is oxidized at the anode to produce \( \mathrm{O}_2 \), \( \mathrm{H}^+ \) ions and electrons: \[ 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4 \mathrm{H}^+ + 4e^- \]
4Step 4: Confirm Anodic Product
From the oxidation of water, the product liberated at the anode is \( \mathrm{O}_2 \).
5Step 5: Answer Verification
Check the multiple-choice options to confirm that \( \mathrm{O}_2 \) is indeed among the given choices.
Key Concepts
Understanding Dilute Sulfuric AcidAnode Reaction in ElectrolysisOxidation in ElectrolysisWater Oxidation in Anode Reactions
Understanding Dilute Sulfuric Acid
Dilute sulfuric acid is a solution where sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) is mixed with water in such a way that its concentration is reduced. In an electrolysis setup, this solution dissociates into ions. Specifically, it separates into hydrogen ions \(\mathrm{H}^{+}\) and sulfate ions \(\mathrm{SO}_{4}^{2-}\). These ions are free to move through the solution and can participate in reactions.
However, during the electrolysis process, not every ion is equally likely to undergo a reaction. The sulfate ions \(\mathrm{SO}_{4}^{2-}\) are usually quite stable and do not get oxidized easily. Instead, the focus often shifts to other available species like water molecules, which play a vital role in the reactions at the electrodes.
However, during the electrolysis process, not every ion is equally likely to undergo a reaction. The sulfate ions \(\mathrm{SO}_{4}^{2-}\) are usually quite stable and do not get oxidized easily. Instead, the focus often shifts to other available species like water molecules, which play a vital role in the reactions at the electrodes.
Anode Reaction in Electrolysis
In any electrolytic cell, the anode is where oxidation reactions take place. The anode is positively charged and attracts anions, or positively charged ions. In the electrolysis of dilute \(\mathrm{H}_{2}\mathrm{SO}_{4}\), determining the reaction at the anode is crucial because it tells us what gets oxidized.
Since sulfate ions \(\mathrm{SO}_{4}^{2-}\) are not easily oxidized under these conditions, the primary reaction at the anode involves water. Water molecules are less stable than sulfate ions and thus more prone to be oxidized. This ultimately leads to the generation of oxygen gas, hydrogen ions, and electrons at the anode.- The key reaction to remember at the anode is the oxidation of water, vital in obtaining our anodic product.
Since sulfate ions \(\mathrm{SO}_{4}^{2-}\) are not easily oxidized under these conditions, the primary reaction at the anode involves water. Water molecules are less stable than sulfate ions and thus more prone to be oxidized. This ultimately leads to the generation of oxygen gas, hydrogen ions, and electrons at the anode.- The key reaction to remember at the anode is the oxidation of water, vital in obtaining our anodic product.
Oxidation in Electrolysis
Oxidation is the process where a substance loses electrons. In the context of electrolysis, specifically with dilute \(\mathrm{H}_{2}\mathrm{SO}_{4}\), oxidation occurs at the anode. The loss of electrons in this setup is associated with water being oxidized rather than sulfate ions.
The oxidation process converts water molecules into oxygen gas. This involves splitting water molecules \(2 \mathrm{H}_2\mathrm{O}\) into oxygen \(\mathrm{O}_2\), releasing hydrogen ions \(\mathrm{H}^+\), and electrons \(4e^-\). The products of this reaction clearly elucidate why oxygen gas appears at the anode during electrolysis of dilute sulfuric acid.
The oxidation process converts water molecules into oxygen gas. This involves splitting water molecules \(2 \mathrm{H}_2\mathrm{O}\) into oxygen \(\mathrm{O}_2\), releasing hydrogen ions \(\mathrm{H}^+\), and electrons \(4e^-\). The products of this reaction clearly elucidate why oxygen gas appears at the anode during electrolysis of dilute sulfuric acid.
- Oxidation reaction: \(2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4 \mathrm{H}^+ + 4e^-\)
- Electrons generated during this reaction contribute to the overall process of electrolysis.
Water Oxidation in Anode Reactions
Water plays a pivotal role during the anode reaction in electrolysis. Given its abundance and reaction potential, water is the molecule that undergoes oxidation in this electrochemical environment.
When water is oxidized, it splits into oxygen gas, hydrogen ions, and electrons. This process can be denoted by the equation:\[2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4 \mathrm{H}^+ + 4e^-\]The reaction is crucial because it's responsible for producing oxygen as the observable product at the anode. Understanding this process allows comprehension of the role water plays in the larger electrolysis mechanism. It's not just about electricity flowing; it's about the intricate reactions that materials undergo.- Water oxidation is an essential concept that aids in understanding how oxygen is liberated at the anode during the electrolysis of dilute sulfuric acid.
When water is oxidized, it splits into oxygen gas, hydrogen ions, and electrons. This process can be denoted by the equation:\[2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{O}_2 + 4 \mathrm{H}^+ + 4e^-\]The reaction is crucial because it's responsible for producing oxygen as the observable product at the anode. Understanding this process allows comprehension of the role water plays in the larger electrolysis mechanism. It's not just about electricity flowing; it's about the intricate reactions that materials undergo.- Water oxidation is an essential concept that aids in understanding how oxygen is liberated at the anode during the electrolysis of dilute sulfuric acid.
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