The amount of electricity that can deposit 108g of silver from silver nitrate solution is

Options

A. 1 ampere
B. 1 coulomb
C. 1 faraday
D. 2 ampere.

Explanation

To determine the amount of electricity that can deposit 108 g of silver from a silver nitrate solution, we can follow these steps:

### Step 1: Understand the Reaction
Silver nitrate (AgNO3) dissociates in solution to give silver ions (Ag⁺) and nitrate ions (NO3⁻). The silver ions gain electrons at the cathode during electrolysis to form solid silver.

### Step 2: Write the Reduction Reaction
The reduction half-reaction for silver ions is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag (s)} \]

### Step 3: Determine the Molar Mass of Silver
The molar mass of silver (Ag) is approximately 108 g/mol.

### Step 4: Calculate the Number of Moles of Silver
To find the number of moles of silver deposited, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For 108 g of silver:
\[ \text{Number of moles of Ag} = \frac{108 \text{ g}}{108 \text{ g/mol}} = 1 \text{ mol} \]

### Step 5: Determine the Equivalent Weight of Silver
Since silver gains one electron to be reduced to metallic silver, its equivalent weight is equal to its molar mass divided by the number of electrons transferred (which is 1 for Ag):
\[ \text{Equivalent weight of Ag} = \frac{108 \text{ g/mol}}{1} = 108 \text{ g/equiv} \]

### Step 6: Calculate the Amount of Charge Required
According to Faraday's laws of electrolysis, the amount of charge (Q) required to deposit one equivalent of a substance is given by:
\[ Q = n \times F \]
where \( n \) is the number of equivalents and \( F \) is Faraday's constant (approximately 96500 C/equiv).

Since we have 1 equivalent of silver:
\[ Q = 1 \text{ equiv} \times 96500 \text{ C/equiv} = 96500 \text{ C} \]

### Conclusion
The amount of electricity required to deposit 108 g of silver from a silver nitrate solution is 96500 C.