Question:
How many moles of platinum will be deposited on the cathode when 0.60 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 0.60 F of electricity is passed through a 1.0 M solution of $Pt^{4+}$?}
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Always divide total Faradays by number of electrons involved in reduction.
Updated On: May 8, 2026
- $0.60$ mol
- $0.15$ mol
- $0.30$ mol
- $0.45$ mol
- $1.0$ mol
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The Correct Option is B
Solution and Explanation
Concept:
Electrolysis follows Faraday’s laws:
• 1 Faraday (F) deposits 1 gram-equivalent of a substance.
• Number of moles deposited: \[ \text{moles} = \frac{\text{charge (in F)}}{\text{number of electrons required}} \]
Step 1: Write half-reaction. \[ Pt^{4+} + 4e^- \rightarrow Pt \] This shows that 4 moles of electrons are required to deposit 1 mole of Pt.
Step 2: Given charge. \[ 0.60 \text{ F} \]
Step 3: Apply Faraday law. \[ \text{moles of Pt} = \frac{0.60}{4} \] \[ = 0.15 \text{ mol} \] Final Answer: \[ \boxed{0.15 \text{ mol}} \]
• 1 Faraday (F) deposits 1 gram-equivalent of a substance.
• Number of moles deposited: \[ \text{moles} = \frac{\text{charge (in F)}}{\text{number of electrons required}} \]
Step 1: Write half-reaction. \[ Pt^{4+} + 4e^- \rightarrow Pt \] This shows that 4 moles of electrons are required to deposit 1 mole of Pt.
Step 2: Given charge. \[ 0.60 \text{ F} \]
Step 3: Apply Faraday law. \[ \text{moles of Pt} = \frac{0.60}{4} \] \[ = 0.15 \text{ mol} \] Final Answer: \[ \boxed{0.15 \text{ mol}} \]
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