Consider the following four electrodes:
`P=Cu^(2+)(0.0001M)//Cu(s)" "Q=Cu^(2+)(0.1M)//Cu(s)`
`R=Cu^(2+)(0.01M)//Cu(s)" "S=Cu^(2+)(0.001M)//Cu(s)`
If the standard electrode potential of `Cu^(2+)//Cu` is +0.34V, the reduction potentials in volts of the above electrodes follow the order:

Options

A. `PgtSgtRgtQ`
B. `SgtRgtQgtP`
C. `RgtSgtQgtP`
D. `QgtRgtSgtP`

Explanation

To solve the problem of determining the order of reduction potentials for the given electrodes, we will use the Nernst equation. The Nernst equation relates the standard electrode potential to the concentration of the reactants and products in an electrochemical reaction.

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation for the reduction of copper ions can be expressed as:
\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]
where \( E^0 \) is the standard electrode potential, \( n \) is the number of electrons transferred (which is 2 for the reduction of \( \text{Cu}^{2+} \)), and \( [\text{Cu}^{2+}] \) is the concentration of copper ions.

2. **Substitute Known Values**:
The standard electrode potential \( E^0 \) for the reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) is given as +0.34 V. Thus, the equation becomes:
\[ E = 0.34 - \frac{0.0591}{2} \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]
This can be simplified to:
\[ E = 0.34 + 0.02955 \log [\text{Cu}^{2+}] \]

3. **Calculate Reduction Potentials for Each Electrode**:
We will calculate the reduction potential for each electrode based on their copper ion concentrations:
- **Electrode P**: \( [\text{Cu}^{2+}] = 0.0001 \, M \)
\[ E_P = 0.34 + 0.02955 \log(0.0001) = 0.34 + 0.02955 \times (-4) = 0.34 - 0.1182 = 0.2218 \, V \]
- **Electrode Q**: \( [\text{Cu}^{2+}] = 0.1 \, M \)
\[ E_Q = 0.34 + 0.02955 \log(0.1) = 0.34 + 0.02955 \times (-1) = 0.34 - 0.02955 = 0.31045 \, V \]
- **Electrode R**: \( [\text{Cu}^{2+}] = 0.01 \, M \)
\[ E_R = 0.34 + 0.02955 \log(0.01) = 0.34 + 0.02955 \times (-2) = 0.34 - 0.0591 = 0.2809 \, V \]
- **Electrode S**: \( [\text{Cu}^{2+}] = 0.001 \, M \)
\[ E_S = 0.34 + 0.02955 \log(0.001) = 0.34 + 0.02955 \times (-3) = 0.34 - 0.08865 = 0.25135 \, V \]

4. **Order the Reduction Potentials**:
Now we can order the reduction potentials calculated:
- \( E_Q = 0.31045 \, V \)
- \( E_R = 0.2809 \, V \)
- \( E_S = 0.25135 \, V \)
- \( E_P = 0.2218 \, V \)

Thus, the order of the reduction potentials from highest to lowest is:
\[ Q > R > S > P \]

### Final Answer:
The reduction potentials in volts of the above electrodes follow the order: **Q > R > S > P**.