An electric current is passed through a silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed `0.108`g more at the end of the electrolysis. The volume of oxygen evolved at STP is:

Options

A. `56 cm^(3)`
B. `550 cm^(3)`
C. `5.6 cm^(3)`
D. `11.2 cm^(3)`

Answer

To solve the problem of finding the volume of oxygen evolved at STP when an electric current is passed through a silver voltameter, we can follow these steps:

### Step 1: Understand the relationship between weight deposited and equivalent weight.
According to Faraday's first law of electrolysis, the weight (W) of a substance deposited at an electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte and the equivalent weight (E) of the substance:
\[ W \propto Q \]
\[ W = n \cdot E \]
where \( n \) is the number of equivalents.

### Step 2: Calculate the equivalent weight of silver (Ag).
The equivalent weight of a substance is calculated using the formula:
\[ E = \frac{M}{n} \]
where \( M \) is the molar mass and \( n \) is the number of electrons involved in the reaction.
For silver (Ag):
- Molar mass of Ag = 108 g/mol
- Since Ag is reduced from Ag⁺ to Ag, it gains 1 electron (n = 1).

Thus, the equivalent weight of Ag is:
\[ E_{Ag} = \frac{108}{1} = 108 \, \text{g/equiv} \]

### Step 3: Calculate the equivalent weight of oxygen (O₂).
For oxygen (O₂) produced during the electrolysis of water:
The reaction is:
\[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
Here, 4 electrons are involved in the formation of 1 mole of O₂, so:
- Molar mass of O₂ = 32 g/mol
- n = 4 (since 4 electrons are lost).

Thus, the equivalent weight of O₂ is:
\[ E_{O_2} = \frac{32}{4} = 8 \, \text{g/equiv} \]

### Step 4: Relate the weight of silver deposited to the weight of oxygen produced.
From the problem, the weight of silver deposited (W₁) is given as 0.108 g. Using the relationship:
\[ \frac{W_{Ag}}{W_{O_2}} = \frac{E_{Ag}}{E_{O_2}} \]
Substituting the known values:
\[ \frac{0.108}{W_{O_2}} = \frac{108}{8} \]

### Step 5: Solve for the weight of oxygen (Wₒ₂).
Cross-multiplying gives:
\[ 0.108 \cdot 8 = 108 \cdot W_{O_2} \]
\[ 0.864 = 108 \cdot W_{O_2} \]
\[ W_{O_2} = \frac{0.864}{108} = 0.008 \, \text{g} \]

### Step 6: Calculate the volume of oxygen at STP.
At STP, 32 g of O₂ occupies 22,400 mL. Therefore, the volume of O₂ produced can be calculated using the unitary method:
\[ \text{Volume of O₂} = \left(\frac{22,400 \, \text{mL}}{32 \, \text{g}}\right) \times 0.008 \, \text{g} \]
\[ \text{Volume of O₂} = 700 \, \text{mL} \]

### Step 7: Final calculation.
To find the volume of oxygen evolved:
\[ \text{Volume of O₂} = \frac{22,400 \times 0.008}{32} = 5.6 \, \text{mL} \]

### Final Answer:
The volume of oxygen evolved at STP is **5.6 mL**.

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