Number of moles of $MnO^{-}_4$ required to oxidise one mole of ferrous oxalate completely in acidic medium will be

Question

The half-life of a 1st order reaction is 1 hr. What is the fraction of the reactant remaining after 3 hours?

Answer 1

The problem requires determining the number of moles of $ \text{MnO}_4^- $ necessary to completely oxidize one mole of ferrous oxalate ($ \text{FeC}_2\text{O}_4 $) in an acidic medium. To solve this, we'll follow these steps:

Understanding the Reaction: In acidic conditions, potassium permanganate ($ \text{MnO}_4^- $) acts as a strong oxidizing agent, which changes to $ \text{Mn}^{2+} $ while gaining electrons. The reaction for this half-cell is:

\text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O
Oxidation of Ferrous Oxalate: Ferrous oxalate contains iron in the $ +2 $ oxidation state, which will be oxidized to $ +3 $ by losing electrons. The half-reaction is:

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\

Moreover, the oxalate ion ($ \text{C}_2\text{O}_4^{2-} $) will be oxidized to carbon dioxide ($ \text{CO}_2 $), the half-reaction being:

\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-\
Combining the Reactions: For one molecule of ferrous oxalate, we have $ \text{Fe}^{2+} $ and $ \text{C}_2\text{O}_4^{2-} $.
- Oxidation of $ \text{Fe}^{2+} $: 1 electron
- Oxidation of $ \text{C}_2\text{O}_4^{2-} $: 2 electrons (because each oxalate releases 2 electrons)
Total electrons needed for complete oxidation of $ \text{FeC}_2\text{O}_4 $ = 1 + 2 = 3 electrons.
Stoichiometry and Moles of $ \text{MnO}_4^- $: From the manganate half-equation, one $ \text{MnO}_4^- $ requires 5 electrons to convert to $ \text{Mn}^{2+} $.

Therefore, the moles of $ \text{MnO}_4^- $ needed:

\frac{3 \text{ electrons}}{5 \text{ electrons per } \text{MnO}_4^-} = 0.6 \text{ mol} \text{ of } \text{MnO}_4^-\

Therefore, to completely oxidize one mole of ferrous oxalate in acidic medium, 0.6 moles of $ \text{MnO}_4^- $ are required. Thus, the correct answer is 0.6 mol.

Answer 2

The problem requires determining the number of moles of $ \text{MnO}_4^- $ necessary to completely oxidize one mole of ferrous oxalate ($ \text{FeC}_2\text{O}_4 $) in an acidic medium. To solve this, we'll follow these steps:

Understanding the Reaction: In acidic conditions, potassium permanganate ($ \text{MnO}_4^- $) acts as a strong oxidizing agent, which changes to $ \text{Mn}^{2+} $ while gaining electrons. The reaction for this half-cell is:

\text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O
Oxidation of Ferrous Oxalate: Ferrous oxalate contains iron in the $ +2 $ oxidation state, which will be oxidized to $ +3 $ by losing electrons. The half-reaction is:

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\

Moreover, the oxalate ion ($ \text{C}_2\text{O}_4^{2-} $) will be oxidized to carbon dioxide ($ \text{CO}_2 $), the half-reaction being:

\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-\
Combining the Reactions: For one molecule of ferrous oxalate, we have $ \text{Fe}^{2+} $ and $ \text{C}_2\text{O}_4^{2-} $.
- Oxidation of $ \text{Fe}^{2+} $: 1 electron
- Oxidation of $ \text{C}_2\text{O}_4^{2-} $: 2 electrons (because each oxalate releases 2 electrons)
Total electrons needed for complete oxidation of $ \text{FeC}_2\text{O}_4 $ = 1 + 2 = 3 electrons.
Stoichiometry and Moles of $ \text{MnO}_4^- $: From the manganate half-equation, one $ \text{MnO}_4^- $ requires 5 electrons to convert to $ \text{Mn}^{2+} $.

Therefore, the moles of $ \text{MnO}_4^- $ needed:

\frac{3 \text{ electrons}}{5 \text{ electrons per } \text{MnO}_4^-} = 0.6 \text{ mol} \text{ of } \text{MnO}_4^-\

Therefore, to completely oxidize one mole of ferrous oxalate in acidic medium, 0.6 moles of $ \text{MnO}_4^- $ are required. Thus, the correct answer is 0.6 mol.

Number of moles of $MnO^{-}_4$ required to oxidise one mole of ferrous oxalate completely in acidic medium will be

The Correct Option is A

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