The electrode potential at 298 K for `Zn//Zn^(2+)` electrode in which the activity of zinc ions is `0.001` M and `E^(o) ""_(Zn//Zn^(2+))` is `0.74` volts will be :

Options

A. `0.38` volts
B. `0.83` volts
C. `0.40` volts
D. `0.45` volts

Answer

To calculate the electrode potential \( E \) for the \( Zn/Zn^{2+} \) electrode at 298 K with the given conditions, we will use the Nernst equation. Here are the steps to solve the problem:

### Step 1: Write the half-reaction
The half-reaction for the zinc electrode can be written as:
\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \]

### Step 2: Identify the given values
- Standard electrode potential \( E^{\circ} \) for \( Zn/Zn^{2+} \) = 0.74 V
- Activity (concentration) of \( Zn^{2+} \) = 0.001 M
- Number of electrons transferred \( n = 2 \)

### Step 3: Write the Nernst equation
The Nernst equation is given by:
\[ E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{[Zn^{2+}]}{[Zn]} \right) \]
Since \( Zn \) is a solid, its activity is considered to be 1, so the equation simplifies to:
\[ E = E^{\circ} - \frac{0.0591}{n} \log \left( [Zn^{2+}] \right) \]

### Step 4: Substitute the known values into the Nernst equation
Substituting the known values into the equation:
\[ E = 0.74 - \frac{0.0591}{2} \log(0.001) \]

### Step 5: Calculate the logarithm
Calculate \( \log(0.001) \):
\[ \log(0.001) = \log(10^{-3}) = -3 \]

### Step 6: Substitute the logarithm value back into the equation
Now substitute \( \log(0.001) \) back into the Nernst equation:
\[ E = 0.74 - \frac{0.0591}{2} \times (-3) \]

### Step 7: Calculate the term involving \( \frac{0.0591}{2} \)
Calculate \( \frac{0.0591}{2} \):
\[ \frac{0.0591}{2} = 0.02955 \]

### Step 8: Substitute and simplify
Now substitute this value back into the equation:
\[ E = 0.74 + (0.02955 \times 3) \]
\[ E = 0.74 + 0.08865 \]
\[ E = 0.82865 \approx 0.83 \text{ V} \]

### Final Answer
Thus, the electrode potential \( E \) for the \( Zn/Zn^{2+} \) electrode is approximately:
\[ \boxed{0.83 \text{ V}} \]