Solveeit Logo

Question

Chemistry Question on Electrochemistry

The cell, ZnZn2+(1M)Cu2+(1M)Cu(Ecell0=1.10V),Zn\left|Zn^{2+}\left(1M\right)\right|\left|Cu^{2+}\left(1M\right)\right|Cu\left(E^{0}_{cell}=1.10V\right), was allowed to be completely discharged at 298K298 \,K. The relative concentration of Zn2+toCu2+[[Zn2+][Cu2+]]Zn^{2+} to Cu^{2+}\left[\frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]}\right] is

A

antilog (24.08)(24.08)

B

37.337.3

C

1037.310^{37.3}

D

9.65?1049.65 ? 10^4

Answer

1037.310^{37.3}

Explanation

Solution

Ecell=Ecello0.0591nlogQE_{cell}=E^{o}_{cell}-\frac{0.0591}{n}log\,Q Where Q=[Zn2+][Cu2+]Q =\frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]} For complete discharge Ecell=0E_{cell}=0 So Ecello=0.5912log[Zn2+][Cu2+]E^{o}_{cell}=\frac{0.591}{2}log \frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]} [Zn2+][Cu2+]=1037.3\Rightarrow \left|\frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]}\right|=10^{37\,.3}