A current is passed through `500mL` of an aqueous solution of `CaI_(2)`. After sometime, it is observed that 50 millimoles of `I_(2)` have been formed. Which of the following statements is `(` are `)` correct ?

Options

A. The number of faradays of charge passed through the solution is `0.10F`.
B. The volume of dry `H_(2)` at `STP` that has been formed during electrolysis is `1120mL`.
C. The `pH` of the solution is nearly `0.7`
D. The mass of calcium produced is `2.0g`.

Explanation

To solve the problem, we need to analyze the electrochemical reactions occurring in the solution of calcium iodide (CaI₂) when a current is passed through it. We will determine the correctness of the provided statements step by step.

### Step 1: Determine the charge passed through the solution
The formation of iodine gas (I₂) at the anode indicates that iodide ions (I⁻) are being oxidized. The reaction can be represented as:
\[ 2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^- \]

From the problem, we know that 50 millimoles of I₂ are produced. To calculate the number of Faradays (F) of charge passed, we use the stoichiometry of the reaction:
- 1 mole of I₂ requires 2 moles of electrons.
- Therefore, 50 millimoles of I₂ will require:
\[ 50 \text{ mmol I}_2 \times 2 \text{ mmol e}^- / \text{mmol I}_2 = 100 \text{ mmol e}^- \]

Since 1 Faraday corresponds to 1 mole of electrons (approximately 96500 coulombs), the number of Faradays used is:
\[ \text{Charge (in Faradays)} = \frac{100 \text{ mmol}}{1000} = 0.1 \text{ Farad} \]

### Step 2: Calculate the volume of hydrogen gas produced
At the cathode, hydrogen ions (H⁺) are reduced to form hydrogen gas (H₂):
\[ 2 \text{H}^+ + 2 \text{e}^- \rightarrow \text{H}_2 \]

Since 100 millimoles of electrons were used, the moles of hydrogen produced will be:
\[ 100 \text{ mmol e}^- \rightarrow 50 \text{ mmol H}_2 \]

At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters (or 22400 mL). Therefore, the volume of hydrogen produced is:
\[ 50 \text{ mmol H}_2 \times \frac{22400 \text{ mL}}{1000 \text{ mmol}} = 1120 \text{ mL} \]

### Step 3: Determine the pH of the solution
As we produced 50 millimoles of H₂, we consumed 100 millimoles of H⁺ ions. The concentration of H⁺ ions consumed in the 500 mL solution is:
\[ \text{Concentration of H}^+ = \frac{100 \text{ mmol}}{500 \text{ mL}} = 0.2 \text{ M} \]

The concentration of OH⁻ ions remaining in the solution can be calculated using the relationship:
\[ \text{pOH} = -\log[\text{OH}^-] \]

Since the concentration of H⁺ ions consumed is 0.2 M, the concentration of OH⁻ ions will also be 0.2 M. Therefore:
\[ \text{pOH} = -\log(0.2) \approx 0.7 \]
And we know that:
\[ \text{pH} + \text{pOH} = 14 \]
Thus:
\[ \text{pH} = 14 - 0.7 = 13.3 \]

### Step 4: Determine the mass of calcium produced
In this scenario, no calcium (Ca) is produced during the electrolysis of CaI₂. Therefore, the mass of calcium produced is 0 grams.

### Conclusion
- **Statement A**: The number of Faradays of charge passed through the solution is 0.1 Farad. **(Correct)**
- **Statement B**: The volume of dry hydrogen at STP that has been formed during electrolysis is 1120 mL. **(Correct)**
- **Statement C**: The pH of the solution is 0.7. **(Incorrect)**
- **Statement D**: The mass of calcium produced is 2 grams. **(Incorrect)**

### Final Answer
The correct statements are A and B.

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