0.02 equivalent of Ag was deposited in an electrolysis experiment. If same quantity of a electricity is passed through a gold solution, 1.314 g of gold is deposited. Find oxidation state of the gold. (Atomic mass of Au = 197)

Answer

To find the oxidation state of gold (Au) based on the given information, we can follow these steps:

### Step 1: Understand the relationship between charge, equivalence, and oxidation state.
The charge (Q) can be expressed in terms of equivalence (Eq) and oxidation state (n):
\[ Q = \text{Equivalence} \times n \]

### Step 2: Set up the equation for silver (Ag).
Given that 0.02 equivalents of silver were deposited, and knowing that the oxidation state of silver is +1:
\[ Q_{Ag} = 0.02 \, \text{Eq} \times 1 \]

### Step 3: Set up the equation for gold (Au).
For gold, we have:
\[ Q_{Au} = \text{Equivalence} \times n \]
Where the equivalence for gold can be calculated using the formula:
\[ \text{Equivalence} = \frac{\text{Given weight}}{\text{Molecular weight}} \]
Given weight of gold deposited is 1.314 g and its atomic mass is 197 g/mol.

### Step 4: Calculate the equivalence of gold.
Substituting the values into the equivalence formula:
\[ \text{Equivalence}_{Au} = \frac{1.314 \, \text{g}}{197 \, \text{g/mol}} \]

### Step 5: Set the charges equal.
Since the same quantity of electricity is passed through both solutions:
\[ Q_{Ag} = Q_{Au} \]
Thus,
\[ 0.02 \, \text{Eq} = \left( \frac{1.314 \, \text{g}}{197 \, \text{g/mol}} \right) \times n \]

### Step 6: Solve for n (oxidation state of gold).
Rearranging the equation gives:
\[ n = \frac{0.02 \times 197}{\frac{1.314}{197}} \]
Calculating the right side:
\[ n = \frac{0.02 \times 197}{1.314 / 197} \]

### Step 7: Perform the calculations.
Calculating the equivalence for gold:
1.314 g / 197 g/mol = 0.00667 Eq (approximately).
Now substituting back:
\[ n = \frac{0.02 \times 197}{0.00667} \]
Calculating this gives:
\[ n \approx 2.99 \]

### Step 8: Round to the nearest whole number.
The oxidation state of gold can be approximated as +3.

### Final Answer:
The oxidation state of gold (Au) is +3.

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