Question

Identify the \(E_{red}\) value from following, if \(\text{CuSO}_{4}\) solution dipped with Cu rod is diluted to \(10\) times.

• A. More by \(0.02958\) V, than \(E_{red}^{0}\)
• B. Less by \(0.02958\) V, than \(E_{red}^{0}\)
• C. More by \(0.0592\) V, than \(E_{red}^{0}\)
• D. Less by \(0.0592\) V, than \(E_{red}^{0}\)

Hint:

Apply the Nernst equation to the Cu2+/Cu half cell and take the concentration as one tenth.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For \(\text{Cu}^{2+}+2e^{-}\rightarrow\text{Cu}\), the Nernst equation is
\(E_{red}=E_{red}^{0}-\dfrac{0.0592}{2}\log\dfrac{1}{[\text{Cu}^{2+}]}\)

Step 2: Effect of Dilution
If the solution is taken as 1 M and diluted 10 times, \([\text{Cu}^{2+}]=0.1\ \text{M}\).

Step 3: Calculation
\(E_{red}=E_{red}^{0}-0.0296\log 10=E_{red}^{0}-0.0296\ \text{V}\)

Step 4: Compare the options
The value is less by about \(0.02958\ \text{V}\), option (B).

Final Answer:
The reduction potential is lower than the standard value by about 0.0296 V. \[ \boxed{\text{(B) }\text{Less by }0.02958\ \text{V, than }E_{red}^{0}} \]