Question:
How many moles of platinum will be deposited on the cathode when 0.40 F of electricity is passed through a 1.0 M solution of \(Pt^{4+}\)?
How many moles of platinum will be deposited on the cathode when 0.40 F of electricity is passed through a 1.0 M solution of \(Pt^{4+}\)?
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For electrodeposition:
\[
\text{Moles deposited}=\frac{\text{Faradays passed}}{\text{charge on ion}}
\]
Updated On: Apr 29, 2026
- 0.60 mol
- 1.0 mol
- 0.40 mol
- 0.45 mol
- 0.10 mol
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The Correct Option is
Solution and Explanation
Concept:
One Faraday deposits one gram equivalent of a substance. For \(Pt^{4+}\), 4 moles of electrons are needed to deposit 1 mole of Pt.
Step 1: Write cathode reaction.
\[ Pt^{4+}+4e^- \rightarrow Pt \]
Step 2: Use given electricity.
\[ 0.40F = 0.40\ \text{mol electrons} \]
Step 3: Calculate moles of Pt.
Since 4 mol electrons deposit 1 mol Pt: \[ \text{Moles of Pt}=\frac{0.40}{4}=0.10 \]
Step 1: Write cathode reaction.
\[ Pt^{4+}+4e^- \rightarrow Pt \]
Step 2: Use given electricity.
\[ 0.40F = 0.40\ \text{mol electrons} \]
Step 3: Calculate moles of Pt.
Since 4 mol electrons deposit 1 mol Pt: \[ \text{Moles of Pt}=\frac{0.40}{4}=0.10 \]
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