`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . If a cell is made by taking the two electrodes what is the feasibility of the reaction?

Options

A. Since `E^@` value for the cell will be positive, redox reaction is feasible.
B. Since `E^@` value for the cell will be negative, redox reaction is not feasible.
C. Ni cannot reduce `Ag^+` to Ag hence reaction is not feasible.
D. Ag can reduce `Ni^(2+)` to Ni hence reaction is feasible.

Explanation

To determine the feasibility of the reaction between the nickel and silver electrodes, we can follow these steps:

### Step 1: Identify the Standard Reduction Potentials
We are given the standard reduction potentials:
- For the half-reaction \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \), \( E^\circ = -0.25 \, \text{V} \)
- For the half-reaction \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \), \( E^\circ = +0.80 \, \text{V} \)

### Step 2: Determine the Anode and Cathode
In an electrochemical cell:
- The half-reaction with the higher reduction potential acts as the cathode (reduction occurs here).
- The half-reaction with the lower reduction potential acts as the anode (oxidation occurs here).

From the given values:
- \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \) (cathode, \( E^\circ = +0.80 \, \text{V} \))
- \( \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \) (anode, \( E^\circ = -0.25 \, \text{V} \))

### Step 3: Write the Half-Reactions
At the anode (oxidation):
\[ \text{Ni} \rightarrow \text{Ni}^{2+} + 2e^- \]

At the cathode (reduction):
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

### Step 4: Balance the Electrons
To balance the electrons transferred in the half-reactions, we need to multiply the cathode reaction by 2:
\[ 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \]

### Step 5: Write the Overall Cell Reaction
Now we can combine the half-reactions:
\[ \text{Ni} + 2\text{Ag}^+ \rightarrow \text{Ni}^{2+} + 2\text{Ag} \]

### Step 6: Calculate the Standard Cell Potential
The standard cell potential \( E^\circ_{\text{cell}} \) is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = (+0.80 \, \text{V}) - (-0.25 \, \text{V}) = +0.80 \, \text{V} + 0.25 \, \text{V} = +1.05 \, \text{V} \]

### Step 7: Determine the Feasibility of the Reaction
Since the standard cell potential \( E^\circ_{\text{cell}} \) is positive (+1.05 V), the reaction is spontaneous. According to the relationship between Gibbs free energy and cell potential:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
A positive \( E^\circ_{\text{cell}} \) indicates that \( \Delta G^\circ \) is negative, confirming that the reaction is feasible.

### Conclusion
The reaction between nickel and silver ions is feasible because the standard cell potential is positive.

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