The cell,Zn | Zn^{2+} ,(1 M) || Cu^{2+} (1 M) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) When the cell is completely discharged,the cell potential $E_{cell} = 0 \ V.$The Nernst equation for the cell reaction is given by:$E_{cell} = E_{

Vedclass · Accepted Answer

$10^{37.3}$

Vedclass · Answer

(D) When the cell is completely discharged,the cell potential $E_{cell} = 0 \ V.$The Nernst equation for the cell reaction is given by:$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} ight)$Here,$n = 2$ (number of electrons transferred) and $E_{cell}^{\circ} = 1.10 \ V.$Substituting the values:$0 = 1.10 - \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} ight)$Rearranging the equation:$\log \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} ight) = \frac{1.10 	imes 2}{0.059} \approx 37.288 \approx 37.3$Therefore,$\frac{[Zn^{2+}]}{[Cu^{2+}]} = 10^{37.3}$.

The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is

Similar Questions

The reduction potential of a hydrogen half-cell will be positive if:

The reduction potential of a hydrogen half-cell will be negative if:

If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

The standard cell potential of a cell reaction involving the transfer of $2$ electrons at $25^{\circ}C$ is $0.295 \ V$. What is its equilibrium constant at $25^{\circ}C$?

$E_{1}$,$E_{2}$ and $E_{3}$ are the emfs of the following three galvanic cells respectively.

Vedclass Test Series

Exam Paper Generator

Online Exam Module