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

Answer

To determine the amount of electricity that can deposit 108 grams 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 reduction of silver ions to metallic silver can be represented by the following half-reaction:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

### Step 2: Calculate Molar Mass of Silver
The molar mass of silver (Ag) is approximately 108 g/mol. This means that 1 mole of silver weighs 108 grams.

### Step 3: Determine Moles of Silver
Since we want to deposit 108 grams of silver, we can calculate the number of moles of silver:
\[ \text{Moles of Ag} = \frac{\text{Mass of Ag}}{\text{Molar Mass of Ag}} = \frac{108 \text{ g}}{108 \text{ g/mol}} = 1 \text{ mol} \]

### Step 4: Relate Moles of Silver to Moles of Electrons
From the half-reaction, we see that 1 mole of silver (Ag) requires 1 mole of electrons (e⁻) for its reduction. Therefore, to deposit 1 mole of silver, we need 1 mole of electrons.

### Step 5: Calculate Charge Required
The charge of 1 mole of electrons is known as 1 Faraday (F), which is approximately 96500 coulombs. Therefore, the charge required to deposit 1 mole of silver is:
\[ \text{Charge} = 1 \text{ F} = 96500 \text{ C} \]

### Conclusion
Thus, the amount of electricity that can deposit 108 grams of silver from silver nitrate solution is **1 Faraday**.