`0.1F` charge is supplied to a solution of `Ni(NO_(3))_(2)`. Then the amount of `Ni` deposited (in mol) at the cathode will be:

Options

A. `0.05`
B. `1.0`
C. `0.5`
D. `0.10`

Answer

To solve the problem of how much nickel (Ni) is deposited at the cathode when 0.1 Faraday of charge is supplied to a solution of Ni(NO₃)₂, we can follow these steps:

### Step 1: Understand the Reaction
The dissociation of nickel(II) nitrate in solution can be represented as:
\[ \text{Ni(NO}_3\text{)}_2 \rightarrow \text{Ni}^{2+} + 2 \text{NO}_3^{-} \]

At the cathode, the nickel ions (\(\text{Ni}^{2+}\)) undergo reduction:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]

### Step 2: Determine the Charge Required for Deposition
From the reduction reaction, we see that 2 moles of electrons are required to deposit 1 mole of nickel. This implies:
- 2 Faraday of charge is needed to deposit 1 mole of nickel.

### Step 3: Calculate Moles of Nickel Deposited
Using the relationship established:
- 1 Faraday deposits \( \frac{1}{2} \) moles of nickel.

Now, if we have 0.1 Faraday of charge, we can calculate the moles of nickel deposited using a unitary method:
\[ \text{Moles of Ni deposited} = \left( \frac{1}{2} \text{ moles of Ni} \right) \times 0.1 \text{ Faraday} \]
\[ \text{Moles of Ni deposited} = \frac{0.1}{2} = 0.05 \text{ moles} \]

### Final Answer
The amount of nickel deposited at the cathode is **0.05 moles**.

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