Question

For the cell reaction,
\(\text{Zn}_{(\text{s})} + 2\text{Ag}^+_{(\text{aq})} \longrightarrow \text{Zn}^{+2}_{(\text{aq})} + 2\text{Ag}_{(\text{s})}\)
Cell potential is less than \(\text{E}^\circ_{\text{cell}}\) by \(0.0592\text{ V}\) at \(298\text{ K}\) when

• A. \([\text{Zn}^{+2}] = 1\text{M}\) and \([\text{Ag}^+] = 0.1\text{M}\)
• B. \([\text{Zn}^{+2}] = 1\text{M}\) and \([\text{Ag}^+] = 0.01\text{M}\)
• C. \([\text{Zn}^{+2}] = 0.1\text{M}\) and \([\text{Ag}^+] = 1\text{M}\)
• D. \([\text{Zn}^{+2}] = 0.01\text{M}\) and \([\text{Ag}^+] = 1\text{M}\)

Hint:

Apply the Nernst equation and find the reaction quotient for which the cell potential drops by 0.0592 V.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For this cell with \(n=2\), the Nernst equation is \(E=E^{\circ}-\frac{0.0592}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2}\) at 298 K.

Step 2: Condition
The cell potential is less than \(E^{\circ}\) by 0.0592 V, so \(\frac{0.0592}{2}\log Q=0.0592\), which gives \(\log Q=2\) and \(Q=100\).

Step 3: Test the options
For \([\text{Zn}^{2+}]=1\text{ M}\) and \([\text{Ag}^{+}]=0.1\text{ M}\): \(Q=\frac{1}{(0.1)^2}=100\).

Step 4: Check the others
The other options give \(Q=10^4\), \(0.1\) and \(0.01\), which do not equal 100.

Step 5: Compare the options
Only the first option gives \(Q=100\).

Final Answer:
The condition holds when \([\text{Zn}^{2+}]=1\text{ M}\) and \([\text{Ag}^{+}]=0.1\text{ M}\). \[ \boxed{\text{(A) }[\text{Zn}^{2+}]=1\text{ M and }[\text{Ag}^{+}]=0.1\text{ M}} \]