The amount of electricity which release 2.0g of gold from a gold salt is same as that which dissolves 0.967g of copper anode during the electrolysis of copper sulphate solution. What is the oxidation number of gold in the gold ion ? (At mass of Cu=63.5,Au=197)

Answer

To solve the problem of determining the oxidation number of gold in the gold ion, we can follow these steps:

### Step 1: Understand the relationship between mass, equivalent weight, and the amount of electricity passed
According to Faraday's second law of electrolysis, the mass of a substance deposited or liberated is proportional to its equivalent weight when the same amount of electricity is passed through different electrolytic cells. This can be expressed as:
\[ \frac{W_1}{E_1} = \frac{W_2}{E_2} \]
where:
- \( W_1 \) = mass of gold deposited (2.0 g)
- \( E_1 \) = equivalent weight of gold
- \( W_2 \) = mass of copper dissolved (0.967 g)
- \( E_2 \) = equivalent weight of copper

### Step 2: Calculate the equivalent weight of copper
The equivalent weight is calculated using the formula:
\[ E = \frac{\text{Atomic weight}}{n} \]
where \( n \) is the number of electrons transferred (valency).
For copper (Cu), the atomic weight is 63.5 g/mol and since copper goes from 0 to +2 oxidation state during electrolysis, \( n = 2 \):
\[ E_2 = \frac{63.5}{2} = 31.75 \, \text{g/equiv} \]

### Step 3: Set up the equation using the mass of gold and copper
Now we can substitute the known values into the ratio:
\[ \frac{2.0}{E_1} = \frac{0.967}{31.75} \]

### Step 4: Calculate the right side of the equation
Calculate \( \frac{0.967}{31.75} \):
\[ \frac{0.967}{31.75} \approx 0.0305 \]

### Step 5: Solve for the equivalent weight of gold
Now we can rearrange the equation to solve for \( E_1 \):
\[ E_1 = \frac{2.0}{0.0305} \approx 65.57 \, \text{g/equiv} \]

### Step 6: Relate equivalent weight of gold to its oxidation number
Using the equivalent weight formula for gold:
\[ E_1 = \frac{\text{Atomic weight of gold}}{n} \]
where the atomic weight of gold is 197 g/mol. Thus:
\[ 65.57 = \frac{197}{n} \]

### Step 7: Solve for \( n \)
Rearranging gives:
\[ n = \frac{197}{65.57} \approx 3.01 \]

### Step 8: Determine the oxidation number
Since \( n \) represents the number of electrons transferred, we can conclude that the oxidation number of gold in the gold ion is approximately +3.

### Final Answer
The oxidation number of gold in the gold ion is +3.

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