Question:
\(38.6\) amperes of current is passed for \(100\) seconds through an aqueous \(CuSO_4\) solution using platinum electrodes. The mass of copper consumed from the solution and volume of gas liberated at STP are respectively
Molar mass of \(Cu=63.54\;g\;mol^{-1}\).
\(38.6\) amperes of current is passed for \(100\) seconds through an aqueous \(CuSO_4\) solution using platinum electrodes. The mass of copper consumed from the solution and volume of gas liberated at STP are respectively
Molar mass of \(Cu=63.54\;g\;mol^{-1}\).
Molar mass of \(Cu=63.54\;g\;mol^{-1}\).
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In electrolysis, first calculate charge using \(Q=It\), then convert it into moles of electrons using \(96500\;C=1\) Faraday.
Updated On: Jun 22, 2026
- \(6.37\;g,\;0.448\;L\)
- \(0.63\;g,\;0.224\;L\)
- \(1.27\;g,\;0.224\;L\)
- \(4\;g,\;0.448\;L\)
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The Correct Option is C
Solution and Explanation
Step 1: Calculate total charge passed.
Current is \[ I=38.6\;A \] Time is \[ t=100\;s \] Total charge passed is \[ Q=It \] \[ Q=38.6\times100 \] \[ Q=3860\;C \]
Step 2: Calculate moles of electrons passed.
One Faraday is \[ 96500\;C \] So, moles of electrons are \[ n_e=\frac{3860}{96500} \] \[ n_e=0.04\;mol \]
Step 3: Calculate mass of copper deposited.
At cathode, \[ Cu^{2+}+2e^-\rightarrow Cu \] So, \(2\) moles of electrons deposit \(1\) mole of copper.
Therefore, moles of copper deposited are \[ n_{Cu}=\frac{0.04}{2} \] \[ n_{Cu}=0.02\;mol \] Mass of copper deposited is \[ m=n_{Cu}\times63.54 \] \[ m=0.02\times63.54 \] \[ m=1.2708\;g \] \[ m\approx1.27\;g \]
Step 4: Calculate volume of gas liberated.
With platinum electrodes, oxygen gas is liberated at the anode: \[ 2H_2O\rightarrow O_2+4H^++4e^- \] Thus, \(4\) moles of electrons liberate \(1\) mole of oxygen gas.
So, moles of oxygen gas are \[ n_{O_2}=\frac{0.04}{4} \] \[ n_{O_2}=0.01\;mol \] At STP, \[ 1\;mol \] of gas occupies \[ 22.4\;L \] Therefore, \[ V=0.01\times22.4 \] \[ V=0.224\;L \]
Step 5: Final conclusion.
Hence, the mass of copper consumed from the solution and volume of gas liberated at STP are \[ \boxed{1.27\;g,\;0.224\;L} \]
Current is \[ I=38.6\;A \] Time is \[ t=100\;s \] Total charge passed is \[ Q=It \] \[ Q=38.6\times100 \] \[ Q=3860\;C \]
Step 2: Calculate moles of electrons passed.
One Faraday is \[ 96500\;C \] So, moles of electrons are \[ n_e=\frac{3860}{96500} \] \[ n_e=0.04\;mol \]
Step 3: Calculate mass of copper deposited.
At cathode, \[ Cu^{2+}+2e^-\rightarrow Cu \] So, \(2\) moles of electrons deposit \(1\) mole of copper.
Therefore, moles of copper deposited are \[ n_{Cu}=\frac{0.04}{2} \] \[ n_{Cu}=0.02\;mol \] Mass of copper deposited is \[ m=n_{Cu}\times63.54 \] \[ m=0.02\times63.54 \] \[ m=1.2708\;g \] \[ m\approx1.27\;g \]
Step 4: Calculate volume of gas liberated.
With platinum electrodes, oxygen gas is liberated at the anode: \[ 2H_2O\rightarrow O_2+4H^++4e^- \] Thus, \(4\) moles of electrons liberate \(1\) mole of oxygen gas.
So, moles of oxygen gas are \[ n_{O_2}=\frac{0.04}{4} \] \[ n_{O_2}=0.01\;mol \] At STP, \[ 1\;mol \] of gas occupies \[ 22.4\;L \] Therefore, \[ V=0.01\times22.4 \] \[ V=0.224\;L \]
Step 5: Final conclusion.
Hence, the mass of copper consumed from the solution and volume of gas liberated at STP are \[ \boxed{1.27\;g,\;0.224\;L} \]
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