Zinc granules are added in excess to `500mL` OF `1.0m` nickel nitrate solution at `25^(@)C` until the equilibrium is reached. If the standard reduction potential of `Zn^(2+)|Zn` and `Ni^(2+)|Ni` are `-0.75V` and `-0.24V`, respectively, find out the concentration of `Ni^(2+)` in solution at equilibrium.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the balanced reaction
The reaction between zinc and nickel ions can be written as:
\[ \text{Zn (s)} + \text{Ni}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Ni (s)} \]

### Step 2: Identify initial concentrations
Initially, we have:
- Concentration of Ni²⁺ = 1.0 M (in 500 mL)
- Concentration of Zn²⁺ = 0 M (since no zinc ions are present at the start)

### Step 3: Set up the equilibrium expressions
At equilibrium, let the concentration of Ni²⁺ that remains be \( A \) M. The concentration of Zn²⁺ produced will be \( 1.0 - A \) M since zinc is added in excess.

### Step 4: Write the Nernst equation
The Nernst equation for the cell can be expressed as:
\[ E_{\text{cell}} = E^\circ_{\text{oxidation}} - E^\circ_{\text{reduction}} + \frac{0.059}{n} \log \left( \frac{[\text{Ni}^{2+}]}{[\text{Zn}^{2+}]} \right) \]

Where:
- \( E^\circ_{\text{oxidation}} = -E^\circ_{\text{Zn}^{2+}/\text{Zn}} = 0.75 \, V \)
- \( E^\circ_{\text{reduction}} = -0.24 \, V \)
- \( n = 2 \) (since 2 electrons are transferred)

### Step 5: Substitute values into the Nernst equation
At equilibrium, we set \( E_{\text{cell}} = 0 \):
\[ 0 = 0.75 - 0.24 + \frac{0.059}{2} \log \left( \frac{A}{1.0 - A} \right) \]

### Step 6: Simplify the equation
This simplifies to:
\[ 0 = 0.51 + 0.0295 \log \left( \frac{A}{1.0 - A} \right) \]

### Step 7: Isolate the logarithmic term
Rearranging gives:
\[ \log \left( \frac{A}{1.0 - A} \right) = -\frac{0.51}{0.0295} \]

### Step 8: Calculate the antilogarithm
Calculating the right side:
\[ \log \left( \frac{A}{1.0 - A} \right) = -17.29 \]
Taking the antilogarithm:
\[ \frac{A}{1.0 - A} = 10^{-17.29} \]

### Step 9: Solve for \( A \)
Let \( x = 10^{-17.29} \):
\[ A = x(1.0 - A) \implies A + Ax = x \implies A(1 + x) = x \implies A = \frac{x}{1 + x} \]

Since \( x \) is very small, \( 1 + x \approx 1 \):
\[ A \approx x \approx 5.15 \times 10^{-18} \, \text{M} \]

### Step 10: Calculate the concentration of Ni²⁺
Since the volume of the solution is 500 mL, the concentration of Ni²⁺ at equilibrium is:
\[ [\text{Ni}^{2+}] = A = 5.15 \times 10^{-18} \, \text{M} \]

### Final Answer
The concentration of Ni²⁺ in solution at equilibrium is:
\[ \text{Concentration of Ni}^{2+} = 5.15 \times 10^{-18} \, \text{M} \]

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