`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 find the EMF of the voltaic cell \( Ni|Ni^{2+} (1.0M)||Au^{3+}(1.0M)|Au \), we will follow these steps:

### Step 1: Identify the half-reactions
The half-reactions for the cell are:
- **Anode (oxidation)**: \( Ni \rightarrow Ni^{2+} + 2e^- \)
- **Cathode (reduction)**: \( Au^{3+} + 3e^- \rightarrow Au \)

### Step 2: Determine the standard reduction potentials
From the question, we have:
- \( E^\circ(Ni^{2+}/Ni) = -0.25 \) V (this is the reduction potential for nickel)
- \( E^\circ(Au^{3+}/Au) = 1.50 \) V (this is the reduction potential for gold)

### Step 3: Calculate the standard EMF of the cell
The standard EMF (\( E^\circ_{cell} \)) can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = E^\circ(Au^{3+}/Au) - E^\circ(Ni^{2+}/Ni) \]
\[ E^\circ_{cell} = 1.50 \, \text{V} - (-0.25 \, \text{V}) = 1.50 \, \text{V} + 0.25 \, \text{V} = 1.75 \, \text{V} \]

### Step 4: Calculate the cell EMF under non-standard conditions
Since both concentrations are 1.0 M, we can use the Nernst equation:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \left( \frac{[Ni^{2+}]^3}{[Au^{3+}]^2} \right) \]
Where \( n \) is the number of electrons transferred in the balanced reaction.

### Step 5: Balance the overall reaction
To balance the half-reactions:
- The oxidation of Ni produces 2 electrons.
- The reduction of Au requires 3 electrons.

To balance the electrons, we multiply the nickel reaction by 3 and the gold reaction by 2:
\[ 3Ni \rightarrow 3Ni^{2+} + 6e^- \]
\[ 2Au^{3+} + 6e^- \rightarrow 2Au \]
Thus, the overall balanced reaction is:
\[ 3Ni + 2Au^{3+} \rightarrow 3Ni^{2+} + 2Au \]
Here, \( n = 6 \).

### Step 6: Substitute into the Nernst equation
Since both concentrations are 1.0 M:
\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{6} \log \left( \frac{1^3}{1^2} \right) \]
\[ E_{cell} = 1.75 \, \text{V} - \frac{0.059}{6} \cdot 0 \]
\[ E_{cell} = 1.75 \, \text{V} \]

### Final Answer
The EMF of the voltaic cell \( Ni|Ni^{2+} (1.0M)||Au^{3+}(1.0M)|Au \) is **1.75 V**.

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