The same quantity of electrical charge that deposited 0.583 g of silver was passed through a solution of gold salt and 0.355 g of gold was formed . What is the oxidation state of gold in this salt ?

Explanation

To determine the oxidation state of gold in the gold salt based on the given information, we can follow these steps:

### Step 1: Calculate the charge (Q) using silver
Using Faraday's first law of electrolysis, we know that:

\[ W = Z \times Q \]

Where:
- \( W \) = mass of the substance deposited (in grams)
- \( Z \) = electrochemical equivalent (in g/C)
- \( Q \) = quantity of charge (in coulombs)

For silver (Ag):
- Mass deposited (\( W_{Ag} \)) = 0.583 g
- Atomic weight of silver (\( A_{Ag} \)) = 108 g/mol
- Number of electrons transferred (\( n_{Ag} \)) = 1 (since Ag⁺ + e⁻ → Ag)

The electrochemical equivalent \( Z_{Ag} \) can be calculated as:

\[ Z_{Ag} = \frac{A_{Ag}}{n_{Ag} \times F} = \frac{108 \, \text{g/mol}}{1 \times 96500 \, \text{C/mol}} = \frac{108}{96500} \, \text{g/C} \]

Now, substituting into the equation for silver:

\[ 0.583 = \left(\frac{108}{96500}\right) \times Q \]

Rearranging to find \( Q \):

\[ Q = \frac{0.583 \times 96500}{108} \approx 529.9 \, \text{C} \]

### Step 2: Calculate the electrochemical equivalent for gold
Now, we will use the same charge \( Q \) to find the oxidation state of gold.

For gold (Au):
- Mass deposited (\( W_{Au} \)) = 0.355 g
- Atomic weight of gold (\( A_{Au} \)) = 197 g/mol
- Let \( n_{Au} \) be the number of electrons transferred.

The electrochemical equivalent \( Z_{Au} \) can be calculated as:

\[ Z_{Au} = \frac{A_{Au}}{n_{Au} \times F} = \frac{197}{n_{Au} \times 96500} \, \text{g/C} \]

Using the equation for gold:

\[ 0.355 = \left(\frac{197}{n_{Au} \times 96500}\right) \times Q \]

Substituting \( Q \):

\[ 0.355 = \left(\frac{197}{n_{Au} \times 96500}\right) \times 529.9 \]

Rearranging to solve for \( n_{Au} \):

\[ n_{Au} = \frac{197 \times 529.9}{0.355 \times 96500} \]

Calculating \( n_{Au} \):

\[ n_{Au} \approx 2.99 \approx 3 \]

### Step 3: Determine the oxidation state of gold
The oxidation state of gold in the salt can be determined from the value of \( n_{Au} \). Since \( n_{Au} \) is the number of electrons transferred, and it is approximately 3, we can conclude that the oxidation state of gold in the gold salt is +3.

### Final Answer
The oxidation state of gold in the gold salt is +3.

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