Question

Which from following statement is correct regarding emf of cell for the cell reaction,

\[ \mathrm{Cd}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{aq})}^{+2} \longrightarrow \mathrm{Cd}_{(\mathrm{aq})}^{+2}+\mathrm{Cu}_{(\mathrm{s})} \]

if concentration of \(\mathrm{Cd}^{+2}\) is 10 times greater than concentration of \(\mathrm{Cu}_{(\mathrm{aq})}^{+2}\) at 298 K .

• A. greater than \(\mathrm{E}_{\text{cell }}^{0}\) cell by 0.0592 V
• B. less than \(\mathrm{E}_{\text{cell }}^{0}\) by 0.0592 V
• C. greater than \(\mathrm{E}_{\text{cell }}^{0}\) by 0.0296 V
• D. less than \(\mathrm{E}_{\text{cell }}^{0}\) by 0.0296 V

Hint:

Apply the Nernst equation with \(Q = [\text{Cd}^{2+}]/[\text{Cu}^{2+}] = 10\) and \(n = 2\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The Nernst equation at 298 K is \(E_{\text{cell}} = E^{0}_{\text{cell}} - \dfrac{0.0592}{n}\log Q\).

Step 2: Reaction quotient
\(\text{Cd} + \text{Cu}^{2+} \rightarrow \text{Cd}^{2+} + \text{Cu}\) has \(n = 2\) and \(Q = \dfrac{[\text{Cd}^{2+}]}{[\text{Cu}^{2+}]} = 10\).

Step 3: Calculation
\(E_{\text{cell}} = E^{0}_{\text{cell}} - \dfrac{0.0592}{2}\log 10 = E^{0}_{\text{cell}} - 0.0296\text{ V}\)

Step 4: Compare the options
The emf is lower than the standard value by 0.0296 V.

Final Answer:
The emf is less than the standard emf by 0.0296 V. \[ \boxed{\text{(D) }\text{less than }E^{0}_{\text{cell}}\text{ by 0.0296 V}} \]