The cell $Zn|Zn^{2+}(1M)||Cu^{2+}(1M)|Cu. (E_{cell}^° = 1.10 V)$ was allowed to be completely discharged at $298 K$. The relative concentration of $Zn^{2+}/Cu^{2+}$ is:
Answer & explanation
Correct answer: option 4
The correct answer is Option 4. $10^{37.3}$
The given cell is \(Zn|Zn^{2+}(1M)||Cu^{2+}(1M)|Cu\)
Thus, the reaction may be represented as:
\(Zn + Cu^{2+} \longrightarrow Zn^{2+} + Cu\)
Given,
\(E^o_{cell} = 1.10 V\)
When the cell is completely discharged, then \(E_{cell} = 0\)
\(∴ E_{cell} = E^o_{cell} - \frac{0.059}{2}log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
\(⇒ 0 = 1.1 - \frac{0.059}{2}log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
\(⇒ 1.1 = \frac{0.059}{2}log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
\(⇒ \frac{2 \times 1.1}{0.059}= log\frac{[Zn^{2+}]}{[Cu^{2+}]}\)
\(⇒ log\frac{[Zn^{2+}]}{[Cu^{2+}]} = 37.3\)
\(⇒ \frac{[Zn^{2+}]}{[Cu^{2+}]} = 10^{37.3}\)