3A current was passed through an aqueous solution of an unknown salt of `Pd` for 1hr. `2.977 g` of `Pd^(+n)` was deposited at cathode. Find n.

Explanation

To solve the problem, we need to determine the value of \( n \) in the deposition of palladium (\( Pd^{+n} \)) at the cathode when a current is passed through its aqueous solution. Here’s a step-by-step solution:

### Step 1: Calculate the moles of palladium deposited
We know the mass of palladium deposited is \( 2.977 \, g \) and the molar mass of palladium is \( 106 \, g/mol \).

\[ \text{Moles of } Pd = \frac{\text{mass}}{\text{molar mass}} = \frac{2.977 \, g}{106 \, g/mol} \]

Calculating this gives:

\[ \text{Moles of } Pd = 0.0281 \, mol \]

### Step 2: Relate moles of electrons to moles of palladium
The reaction at the cathode can be represented as:

\[ Pd^{+n} + ne^- \rightarrow Pd \]

From this reaction, we can see that for every mole of \( Pd \) deposited, \( n \) moles of electrons are required. Therefore, the total moles of electrons (\( n \times \text{moles of } Pd \)) can be expressed as:

\[ \text{Moles of electrons} = n \times \text{Moles of } Pd = n \times 0.0281 \]

### Step 3: Calculate the total charge passed
Using the formula for total charge \( Q \):

\[ Q = I \times t \]

Where:
- \( I = 3 \, A \)
- \( t = 1 \, hr = 3600 \, s \)

Calculating the total charge:

\[ Q = 3 \, A \times 3600 \, s = 10800 \, C \]

### Step 4: Relate total charge to moles of electrons
Using Faraday's constant (\( F = 96500 \, C/mol \)), we can express the total charge in terms of moles of electrons:

\[ Q = \text{Moles of electrons} \times F \]

Substituting the expression for moles of electrons:

\[ 10800 = (n \times 0.0281) \times 96500 \]

### Step 5: Solve for \( n \)
Rearranging the equation to solve for \( n \):

\[ n = \frac{10800}{0.0281 \times 96500} \]

Calculating the right side:

\[ n = \frac{10800}{2713.15} \approx 3.98 \]

Rounding this value gives \( n = 4 \).

### Conclusion
Thus, the value of \( n \) is \( 4 \).

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