The cell emf for the cell `Ni(s)abs(Ni^(2+)(1.0M))Au^(3+)(1.0M)| Au(s) (E^(o)` for `Ni^(2+) abs(Ni=-0.25 V,E^(o)" for " Au^(3+))Au=1.50V)` is

Options

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

Answer

To find the cell emf (E_cell) for the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials (E°)
- For Nickel (Ni):
- The half-reaction is:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]
- Given E° for this reaction is:
\[ E°_{\text{Ni}^{2+}/\text{Ni}} = -0.25 \, \text{V} \]

- For Gold (Au):
- The half-reaction is:
\[ \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \]
- Given E° for this reaction is:
\[ E°_{\text{Au}^{3+}/\text{Au}} = +1.50 \, \text{V} \]

### Step 2: Determine the oxidation and reduction reactions
- In this cell, Nickel is oxidized (loses electrons) and Gold is reduced (gains electrons).
- The oxidation half-reaction for Nickel can be written as:
\[ \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \]
- The E° for oxidation is the negative of the reduction potential:
\[ E°_{\text{oxidation Ni}} = +0.25 \, \text{V} \]

### Step 3: Use the cell emf formula
- The formula for the cell emf is:
\[ E_{\text{cell}} = E°_{\text{reduction}} + E°_{\text{oxidation}} \]

### Step 4: Substitute the values into the formula
- Now substituting the values we have:
\[ E_{\text{cell}} = E°_{\text{Au}^{3+}/\text{Au}} + E°_{\text{oxidation Ni}} \]
\[ E_{\text{cell}} = 1.50 \, \text{V} + 0.25 \, \text{V} \]

### Step 5: Calculate the cell emf
- Performing the addition:
\[ E_{\text{cell}} = 1.75 \, \text{V} \]

### Final Answer
The cell emf for the given electrochemical cell is:
\[ \boxed{1.75 \, \text{V}} \]

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