Find the amount of silver liberated at cathode if `0.500 A` of current is passed through `AgNO_3` electrolyte for 1 hour. Atomic weight of silver s `107.9 g mol^(-1)`

Explanation

To find the amount of silver liberated at the cathode when a current of 0.500 A is passed through an AgNO₃ electrolyte for 1 hour, we can use Faraday's laws of electrolysis. Here’s a step-by-step solution:

### Step 1: Understand the Formula
According to Faraday's laws of electrolysis, the mass of the substance liberated at an electrode is given by the formula:
\[ M = Z \cdot I \cdot T \]
where:
- \( M \) = mass of the substance liberated (in grams)
- \( Z \) = electrochemical equivalent (in grams per coulomb)
- \( I \) = current (in amperes)
- \( T \) = time (in seconds)

### Step 2: Calculate the Electrochemical Equivalent (Z)
The electrochemical equivalent \( Z \) can be calculated using the formula:
\[ Z = \frac{\text{Equivalent weight}}{F} \]
where:
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- Equivalent weight = \(\frac{\text{Atomic weight}}{\text{Valency}}\)

For silver (Ag):
- Atomic weight = \( 107.9 \, \text{g/mol} \)
- Valency = 1 (since silver typically forms +1 ions)

Thus, the equivalent weight of silver is:
\[ \text{Equivalent weight} = \frac{107.9}{1} = 107.9 \, \text{g/mol} \]

Now, substituting this into the formula for \( Z \):
\[ Z = \frac{107.9}{96500} \approx 0.001118 \, \text{g/C} \]

### Step 3: Convert Time to Seconds
The time given is 1 hour. We need to convert this into seconds:
\[ T = 1 \, \text{hour} = 3600 \, \text{seconds} \]

### Step 4: Substitute Values into the Formula
Now we can substitute the values into the formula for \( M \):
- Current \( I = 0.500 \, \text{A} \)
- Time \( T = 3600 \, \text{s} \)
- Electrochemical equivalent \( Z \approx 0.001118 \, \text{g/C} \)

So,
\[ M = Z \cdot I \cdot T \]
\[ M = 0.001118 \cdot 0.500 \cdot 3600 \]

### Step 5: Calculate the Mass of Silver Liberated
Now we will perform the calculation:
\[ M = 0.001118 \cdot 0.500 \cdot 3600 \]
\[ M = 0.001118 \cdot 1800 \]
\[ M \approx 2.01 \, \text{g} \]

### Final Answer
The amount of silver liberated at the cathode is approximately **2.01 grams**.

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