The standard electrode potential (reduction) of `Ag^(+)|Ag` is `0.800 V` at `25^(@)C`. Its electrode potential in a solution containing `10^(-3)M` ion of `Ag^(+)` ions is:-

Options

A. `0.623V`
B. `-0.977V`
C. `0.892V`
D. `1.246V`

Explanation

To find the electrode potential of the Ag⁺/Ag system in a solution containing 10⁻³ M Ag⁺ ions, we will use the Nernst equation. Here’s the step-by-step solution:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
where:
- \( E \) = electrode potential under non-standard conditions
- \( E^\circ \) = standard electrode potential
- \( n \) = number of moles of electrons transferred in the half-reaction
- \( Q \) = reaction quotient

### Step 2: Identify the Standard Electrode Potential
From the problem, we know:
- \( E^\circ = 0.800 \, V \)

### Step 3: Determine the Number of Electrons Transferred
For the reduction of Ag⁺ to Ag:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
Here, \( n = 1 \) (one electron is transferred).

### Step 4: Calculate the Reaction Quotient (Q)
In this case, the reaction quotient \( Q \) is given by the concentration of the reactants. Since the product (Ag) is a solid, its activity is considered to be 1. Thus:
\[ Q = \frac{1}{[\text{Ag}^+]} = \frac{1}{10^{-3}} = 10^{3} \]

### Step 5: Substitute Values into the Nernst Equation
Now substituting the values into the Nernst equation:
\[ E = 0.800 \, V - \frac{0.0591}{1} \log(10^{3}) \]

### Step 6: Simplify the Logarithm
Since \( \log(10^{3}) = 3 \):
\[ E = 0.800 \, V - 0.0591 \times 3 \]

### Step 7: Calculate the Final Value
Now, calculate:
\[ E = 0.800 \, V - 0.1773 \]
\[ E = 0.6227 \, V \]

### Step 8: Round the Result
Rounding to three decimal places, we get:
\[ E \approx 0.623 \, V \]

### Final Answer
The electrode potential in a solution containing \( 10^{-3} \, M \) Ag⁺ ions is approximately **0.623 V**.

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