`E^(@)(Ni^(2+)//Ni) =- 0.25` volt, `E^(@)(Au^(3+)//Au) = 1.50` volt. The emf of the voltaic cell `Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au` is:-

Options

A. `1.25` volt
B. `-1.75` volt
C. `1.75` volt
D. `4.0`volt

Explanation

To solve the problem of finding the emf of the voltaic cell \( Ni | Ni^{2+} (1.0M) || Au^{3+} (1.0M) | Au \), we can follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials
The standard reduction potentials given are:
- For nickel:
\[ E^\circ(Ni^{2+}/Ni) = -0.25 \, \text{V} \]
- For gold:
\[ E^\circ(Au^{3+}/Au) = 1.50 \, \text{V} \]

### Step 2: Write the half-reactions
The half-reaction at the anode (oxidation) for nickel is:
\[ Ni \rightarrow Ni^{2+} + 2e^- \]
The half-reaction at the cathode (reduction) for gold is:
\[ Au^{3+} + 3e^- \rightarrow Au \]

### Step 3: Balance the half-reactions
To balance the number of electrons transferred, we need to multiply the nickel half-reaction by 3 and the gold half-reaction by 2:
- Nickel half-reaction (multiplied by 3):
\[ 3Ni \rightarrow 3Ni^{2+} + 6e^- \]
- Gold half-reaction (multiplied by 2):
\[ 2Au^{3+} + 6e^- \rightarrow 2Au \]

### Step 4: Combine the half-reactions
Now, we can add the two balanced half-reactions:
\[ 3Ni + 2Au^{3+} \rightarrow 3Ni^{2+} + 2Au \]

### Step 5: Calculate the standard cell potential \( E^\circ_{cell} \)
Using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 1.50 \, \text{V} - (-0.25 \, \text{V}) = 1.50 \, \text{V} + 0.25 \, \text{V} = 1.75 \, \text{V} \]

### Step 6: Apply the Nernst equation
Since the concentrations of both ions are 1.0 M, we can use the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \left( \frac{[products]}{[reactants]} \right) \]
Where \( n = 6 \) (the number of electrons transferred).

Substituting the values:
\[ E_{cell} = 1.75 \, \text{V} - \frac{0.059}{6} \log \left( \frac{(Ni^{2+})^3}{(Au^{3+})^2} \right) \]
Since both concentrations are 1.0 M:
\[ E_{cell} = 1.75 \, \text{V} - \frac{0.059}{6} \log(1) = 1.75 \, \text{V} - 0 = 1.75 \, \text{V} \]

### Final Answer
The emf of the voltaic cell is:
\[ \boxed{1.75 \, \text{V}} \]