Emf of the cell
`Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M)` Au will be
`E_(Ni//Ni(2+))^@ = 0.5=25. E_(Au//Au^(3+))^@ = 1.5 V`.

Options

A. ` 1. 75 V`
B. ` +1. 7795 V`
C. ` = 0 . 775 V`
D. `- 1.7795 V`

Explanation

To calculate the EMF of the cell represented by the notation `Ni | Ni^(2+) (0.1 M) | Au^(3+) (1.0 M)`, we will follow these steps:

### Step 1: Identify the standard reduction potentials
We have the following standard reduction potentials:
- For the Ni/Ni²⁺ half-reaction:
\[ E^\circ_{\text{Ni/Ni}^{2+}} = 0.25 \, \text{V} \]
- For the Au/Au³⁺ half-reaction:
\[ E^\circ_{\text{Au/Au}^{3+}} = 1.5 \, \text{V} \]

### Step 2: Write the balanced cell reaction
The balanced overall reaction for the cell is:
\[ 3 \, \text{Ni} + 2 \, \text{Au}^{3+} \rightarrow 3 \, \text{Ni}^{2+} + 2 \, \text{Au} \]

### Step 3: Determine the number of electrons transferred
From the balanced equation, we see that 6 electrons are transferred (3 Ni atoms oxidized to Ni²⁺ and 2 Au³⁺ reduced to Au):
\[ n = 6 \]

### Step 4: Calculate the standard EMF of the cell
The standard EMF of the cell can be calculated using:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, Au is the cathode (reduction) and Ni is the anode (oxidation):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Au/Au}^{3+}} - E^\circ_{\text{Ni/Ni}^{2+}} \]
\[ E^\circ_{\text{cell}} = 1.5 \, \text{V} - 0.25 \, \text{V} = 1.25 \, \text{V} \]

### Step 5: Apply the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Ni}^{2+}]^3}{[\text{Au}^{3+}]^2} \right) \]

Substituting the values:
- \([Ni^{2+}] = 0.1 \, \text{M}\)
- \([Au^{3+}] = 1.0 \, \text{M}\)

The Nernst equation becomes:
\[ E_{\text{cell}} = 1.25 \, \text{V} - \frac{0.0591}{6} \log \left( \frac{(0.1)^3}{(1.0)^2} \right) \]

### Step 6: Calculate the logarithmic term
Calculating the logarithm:
\[ \log \left( \frac{(0.1)^3}{(1.0)^2} \right) = \log(0.001) = -3 \]

### Step 7: Substitute the logarithmic value back into the Nernst equation
Now substituting this back:
\[ E_{\text{cell}} = 1.25 \, \text{V} - \frac{0.0591}{6} \times (-3) \]
\[ E_{\text{cell}} = 1.25 \, \text{V} + \frac{0.0591 \times 3}{6} \]
\[ E_{\text{cell}} = 1.25 \, \text{V} + 0.02955 \, \text{V} \]
\[ E_{\text{cell}} = 1.27955 \, \text{V} \]

### Final Result
Thus, the EMF of the cell is approximately:
\[ E_{\text{cell}} \approx 1.28 \, \text{V} \]