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 solve the problem, we need to find the oxidation state of gold (Au) based on the information given about the electrolysis experiment. Here’s the step-by-step solution:

### Step 1: Understand the concept of equivalents
In electrolysis, the amount of substance deposited is directly proportional to the quantity of electricity passed. The equivalent of a substance is defined as the mass of the substance that will combine with or displace one mole of hydrogen or 1/96.485 moles of oxygen.

### Step 2: Determine the equivalent of silver (Ag)
From the problem, we know that 0.02 equivalents of Ag were deposited. The half-reaction for silver is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
This means that 1 equivalent of Ag corresponds to 1 mole of electrons (n = 1). Therefore, 0.02 equivalents of Ag corresponds to:
\[ \text{Mass of Ag} = \text{Equivalent} \times \text{Equivalent weight of Ag} \]
The equivalent weight of Ag (atomic mass = 107.87 g/mol) is:
\[ \text{Equivalent weight of Ag} = \frac{107.87}{1} = 107.87 \, \text{g/equiv} \]
Thus, the mass of Ag deposited is:
\[ \text{Mass of Ag} = 0.02 \times 107.87 = 2.1574 \, \text{g} \]

### Step 3: Determine the equivalent of gold (Au)
When the same quantity of electricity is passed through a gold solution, 1.314 g of gold is deposited. The atomic mass of gold (Au) is 197 g/mol. We need to find the number of equivalents of gold deposited:
\[ \text{Equivalent weight of Au} = \frac{197}{n} \]
Where \( n \) is the number of electrons transferred per atom of gold. The mass of gold deposited can be expressed in terms of equivalents:
\[ \text{Mass of Au} = \text{Equivalent} \times \text{Equivalent weight of Au} \]
Let \( x \) be the number of equivalents of gold deposited:
\[ 1.314 = x \times \frac{197}{n} \]

### Step 4: Set up the equation based on equivalence
Since the same quantity of electricity was passed through both solutions, the equivalents of silver and gold are equal:
\[ 0.02 = x \]
Thus, substituting \( x \):
\[ 1.314 = 0.02 \times \frac{197}{n} \]

### Step 5: Solve for \( n \)
Rearranging the equation to solve for \( n \):
\[ n = \frac{0.02 \times 197}{1.314} \]
Calculating this gives:
\[ n = \frac{3.94}{1.314} \approx 3 \]

### Step 6: Determine the oxidation state of gold
Since \( n = 3 \), this indicates that gold is in the +3 oxidation state when it is deposited:
\[ \text{Oxidation state of Au} = +3 \]

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

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