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Consider the following half-equations, which are not in standard electrode potential order - VCE - SSCE Chemistry - Question 30 - 2022 - Paper 1
Question 30
Consider the following half-equations, which are not in standard electrode potential order. HCrO4^{-} (aq) + 7H^{+} (aq) + 3e^{-} ightarrow Cr^{3+} (aq) + 4H2O (l)... show full transcript
Worked Solution & Example Answer:Consider the following half-equations, which are not in standard electrode potential order - VCE - SSCE Chemistry - Question 30 - 2022 - Paper 1
Step 1
Identify the half-cell reactions
Answer
We will first identify and write down the half-cell reactions from the question:
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For HCrO4^{-} (aq):
- Reduction: HCrO4^{-} + 7H^{+} + 3e^{-} → Cr^{3+} + 4H2O
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For HBrO3 (aq):
- Reduction: HBrO3 + H^{+} + rac{1}{2} Br2 → H2O
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For I^{-} (aq):
- Oxidation: 2I^{-} + 12H^{+} + 10e^{-} → 6H2O
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For BrO3^{-} (aq):
- Reduction: BrO3^{-} + 6H^{+} + 6e^{-} → Br^{-} + 3H2O.
Step 2
Analyze galvanic cell combinations
Answer
Next, we will analyze the provided combinations:
- Option A: HBRO(aq)/BrO3(aq) | HCrO4(aq)/Cr^{3+}(aq)
- Option B: HCrO4(aq)/Cr^{3+}(aq) | BrO3(aq)/Br^{-}(aq)
- Option C: BrO3(aq)/Br^{-}(aq) | HBrO(aq)/I2(aq)
- Option D: HBrO(aq)/Br(aq) | HBrO(aq)/I3(aq)
Now, we will evaluate which combination yields the highest potential difference by considering the strengths of the reductions. HCrO4^{-} has a higher reduction potential compared to HBrO3.
Step 3