Question

The number of moles of \(H_{2}\) gas liberated at cathode, when 10 mA current is passed through dilute NaCl for \(19.3 \times 10^{4}\) seconds is (\(F=96500 C mol^{-1}\))

• A. 0.50
• B. 0.02
• C. 0.01
• D. 0.15

Hint:

Always double-check the cathodic reaction (reduction) and the specific stoichiometric coefficient for electrons to correctly determine \(n\).

Answer 1

The Correct Option is C

Step 1: Recall Faraday's law.
The amount of product equals charge divided by ($n$ times Faraday): $\text{moles}=\frac{Q}{nF}$.
Step 2: Find the charge.
Current $=10$ mA $=0.01$ A and time $=19.3\times10^4$ s. \[ Q=0.01\times193000=1930\ \text{C} \]
Step 3: Write the cathode reaction.
In dilute NaCl, water is reduced: $2H_2O+2e^-\rightarrow H_2+2OH^-$. So 2 electrons make 1 molecule of $H_2$, giving $n=2$.
Step 4: Put values in the formula.
\[ \text{moles of }H_2=\frac{1930}{2\times96500} \]
Step 5: Do the arithmetic.
\[ =\frac{1930}{193000}=0.01\ \text{mol} \]
Step 6: State the answer.
\[ \boxed{0.01\ \text{mol of }H_2} \]