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 solve the problem, we need to determine the cell emf (E_cell) for the given electrochemical cell reaction. The cell is represented as:

\[ \text{Ni(s)} | \text{Ni}^{2+}(1.0M) || \text{Au}^{3+}(1.0M) | \text{Au(s)} \]

Given standard reduction potentials:
- \( E^\circ \) for \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \) is -0.25 V
- \( E^\circ \) for \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \) is +1.50 V

### Step 1: Identify the Anode and Cathode
- The half-reaction with a lower reduction potential acts as the anode (oxidation).
- The half-reaction with a higher reduction potential acts as the cathode (reduction).

Here, since \( E^\circ \) for Au is higher than that for Ni, we have:
- Anode: \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \) (oxidation)
- Cathode: \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \) (reduction)

### Step 2: Calculate the Standard Cell Potential (E°_cell)
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 1.50 \, \text{V} - (-0.25 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 1.50 \, \text{V} + 0.25 \, \text{V} \]
\[ E^\circ_{\text{cell}} = 1.75 \, \text{V} \]

### Step 3: Calculate the Cell Potential (E_cell) using the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

Where:
- \( n \) is the number of moles of electrons transferred in the balanced equation.

### Step 4: Balance the Overall Reaction
To balance the overall reaction, we need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. The half-reactions are:
1. \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \) (2 electrons)
2. \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \) (3 electrons)

To balance, we can multiply the nickel half-reaction by 3 and the gold half-reaction by 2:
- \( 3 \text{Ni} \rightarrow 3 \text{Ni}^{2+} + 6e^- \)
- \( 2 \text{Au}^{3+} + 6e^- \rightarrow 2 \text{Au} \)

### Step 5: Substitute into the Nernst Equation
Now we have \( n = 6 \) (total electrons transferred). The concentrations of both ions are given as 1.0 M, so:
\[ E_{\text{cell}} = 1.75 \, \text{V} - \frac{0.0591}{6} \log \left( \frac{[2 \text{Au}]}{[3 \text{Ni}^{2+}]^3} \right) \]

Since both concentrations are 1.0 M, we have:
\[ E_{\text{cell}} = 1.75 \, \text{V} - \frac{0.0591}{6} \log(1) \]
Since \( \log(1) = 0 \):
\[ E_{\text{cell}} = 1.75 \, \text{V} - 0 \]
\[ E_{\text{cell}} = 1.75 \, \text{V} \]

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