Question

If 5 amperes of current is passed for 193 seconds through a solution containing copper salt, \(0.32\ g\) of copper is deposited. What is the oxidation state of Cu in the salt?

• A. \(+2\)
• B. \(+1\)
• C. \(+3\)
• D. \(+\dfrac{3}{2}\)

Hint:

In electrolysis, \[ Q=It \] and \[ 1\ \text{Faraday}=96500\ C \] Always calculate the moles of electrons first and then relate them to the electrode reaction to determine the oxidation state.

Answer 1

The Correct Option is A

Step 1: Calculate the total charge passed.
Using \[ Q=It \] where \[ I=5\ A \] and \[ t=193\ s \] Therefore, \[ Q=5\times 193 \] \[ Q=965\ C \]

Step 2: Calculate the number of moles of electrons passed.
One Faraday is \[ 96500\ C \] and corresponds to one mole of electrons.
Hence, \[ \text{Moles of electrons} = \frac{965}{96500} \] \[ =0.01 \]

Step 3: Apply Faraday's law of electrolysis.
Let the oxidation state of copper in the salt be \(n\). The cathode reaction is \[ Cu^{n+}+ne^- \rightarrow Cu \] Thus, \(n\) moles of electrons deposit one mole of copper.
Given mass of copper deposited, \[ m=0.32\ g \] Atomic mass of copper, \[ M=64\ g\,mol^{-1} \] Therefore, moles of copper deposited are \[ \frac{0.32}{64} \] \[ =0.005 \]

Step 4: Determine the oxidation state.
Using \[ \text{Moles of electrons} = n\times \text{Moles of Cu deposited} \] \[ 0.01=n\times 0.005 \] \[ n=\frac{0.01}{0.005} \] \[ n=2 \]

Step 5: Final conclusion.
Therefore, the oxidation state of copper in the salt is \[ \boxed{+2} \] Hence, option (1) is correct.