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Step-by-Step Solution
Step 1: Identify the Reaction at the Cathode
During the electrolysis of acidified water, hydrogen gas is released at the cathode according to the half-reaction:
$2\,H^+ + 2\,e^- \to H_2.$
Step 2: Calculate the Number of Moles of Hydrogen Gas Collected
At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies $22400\,\text{mL}$. Since $112\,\text{mL}$ of $H_2$ was collected:
$\text{Moles of }H_2 = \frac{112\,\text{mL}}{22400\,\text{mL/mol}} = 0.005\,\text{mol}.$
Step 3: Relate the Mass of Hydrogen Gas to the Faraday's Laws
The molar mass of $H_2$ is $2\,\text{g/mol}$. Hence the mass of $H_2$ gas collected is:
$\text{Mass of }H_2 = 0.005\,\text{mol} \times 2\,\text{g/mol} = 0.01\,\text{g}.$
Step 4: Use Faraday’s Law to Find the Electric Charge
According to Faraday's first law of electrolysis, the mass of a substance deposited or liberated at an electrode (w) is directly proportional to the total electric charge (Q) passed:
$w = \frac{E \times I \times t}{96500},$
where:
$w$ = mass of the substance (in grams),
$E$ = equivalent mass (for $H_2$, equivalence is effectively $1\,\text{g/equivalent}$ since $2\,\text{g}$ of $H_2$ requires $2$ moles of electrons),
$I$ = current (in amperes),
$t$ = time (in seconds), and
$96500\,\text{C/mol}$ is the Faraday constant.
Step 5: Substitute the Known Values and Solve for Current
We have $w = 0.01\,\text{g}$, $t = 965\,\text{s}$, and $E = 1\,\text{g/equivalent}$. Plugging these values in:
$0.01 = \frac{1 \times I \times 965}{96500}.$
Rearranging to solve for $I$:
$I = \frac{0.01 \times 96500}{965} = 1\,\text{A}.
Final Answer
The current passing through the acidified water is 1 A.
