Question:
If the total electricity required to deposit 1 mole of a metal M is equal to that of 10.7 amperes of current for 10 hours, the equivalent weight of the metal is (atomic weight = M u):
If the total electricity required to deposit 1 mole of a metal M is equal to that of 10.7 amperes of current for 10 hours, the equivalent weight of the metal is (atomic weight = M u):
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Use the formula \(E = \frac{It}{96500 \times n}\) to determine equivalent weight from electrolysis data, where \(n\) is the valency of the metal.
Updated On: Jun 26, 2026
- M
- M/2
- M/3
- M/4
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The Correct Option is D
Solution and Explanation
Step 1: Recall the relation between electricity and equivalent weight.
The equivalent weight of a metal deposited by electrolysis is given by \[ E = \frac{It}{96500 \times n} \] where \(I\) is current in amperes, \(t\) is time in seconds, \(n\) is the number of electrons per atom, and \(96500\) C is 1 Faraday.
Step 2: Calculate total charge.
\[ Q = I \times t = 10.7 \, \text{A} \times 10 \, \text{hours} \times 3600 \, \text{s/hour} = 385,200 \, \text{C} \]
Step 3: Relate to equivalent weight.
1 mole requires 2 Faradays for a divalent metal: \[ E = \frac{M}{4} \quad (\text{since } 2 \text{ electrons, 2 Faradays}) \]
Step 4: Conclusion.
Hence, the equivalent weight of the metal is \[ \boxed{M/4}. \]
The equivalent weight of a metal deposited by electrolysis is given by \[ E = \frac{It}{96500 \times n} \] where \(I\) is current in amperes, \(t\) is time in seconds, \(n\) is the number of electrons per atom, and \(96500\) C is 1 Faraday.
Step 2: Calculate total charge.
\[ Q = I \times t = 10.7 \, \text{A} \times 10 \, \text{hours} \times 3600 \, \text{s/hour} = 385,200 \, \text{C} \]
Step 3: Relate to equivalent weight.
1 mole requires 2 Faradays for a divalent metal: \[ E = \frac{M}{4} \quad (\text{since } 2 \text{ electrons, 2 Faradays}) \]
Step 4: Conclusion.
Hence, the equivalent weight of the metal is \[ \boxed{M/4}. \]
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