Question:
Three electrolytic cells P, Q and R containing aqueous solutions of \(Al(NO_3)_3\), \(AgNO_3\) and \(CuSO_4\) respectively are connected in series. A current of 1.5 A was passed until 1.45 g of Ag was deposited. Find the masses of Al and Cu deposited.
Three electrolytic cells P, Q and R containing aqueous solutions of \(Al(NO_3)_3\), \(AgNO_3\) and \(CuSO_4\) respectively are connected in series. A current of 1.5 A was passed until 1.45 g of Ag was deposited. Find the masses of Al and Cu deposited.
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In series connection, the same quantity of electricity passes through all cells.
Updated On: Jun 17, 2026
- 0.426, 0.121
- 0.363, 0.852
- 0.121, 0.852
- 0.121, 0.426
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The Correct Option is C
Solution and Explanation
Concept:
According to Faraday's Second Law:
\[
m\propto \frac{M}{n}
\]
where \(M\) is molar mass and \(n\) is number of electrons involved.
Step 1: Calculate equivalent masses.
For silver: \[ E_{Ag}=\frac{108}{1}=108 \] For aluminium: \[ E_{Al}=\frac{27}{3}=9 \] For copper: \[ E_{Cu}=\frac{63.5}{2}=31.75 \]
Step 2: Use Faraday's law.
\[ \frac{m_{Al}}{m_{Ag}} = \frac{9}{108} \] \[ m_{Al} = 1.45\times\frac{9}{108} \] \[ m_{Al}=0.121g \] \[ m_{Cu} = 1.45\times\frac{31.75}{108} \] \[ m_{Cu}=0.426g \]
Step 3: Final answer.
\[ \boxed{0.121g,\;0.426g} \]
Step 1: Calculate equivalent masses.
For silver: \[ E_{Ag}=\frac{108}{1}=108 \] For aluminium: \[ E_{Al}=\frac{27}{3}=9 \] For copper: \[ E_{Cu}=\frac{63.5}{2}=31.75 \]
Step 2: Use Faraday's law.
\[ \frac{m_{Al}}{m_{Ag}} = \frac{9}{108} \] \[ m_{Al} = 1.45\times\frac{9}{108} \] \[ m_{Al}=0.121g \] \[ m_{Cu} = 1.45\times\frac{31.75}{108} \] \[ m_{Cu}=0.426g \]
Step 3: Final answer.
\[ \boxed{0.121g,\;0.426g} \]
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