Question

$E^\circ $ for the cell, $Zn|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu $ is 1.10 V at $25^\circ C. $ The equilibrium constant for the reaction, $Zn(s)+Cu^{2+}(aq) \rightleftharpoons Cu(s)+Zn^{2+}(aq) $ is of the order

• A. $10^{37} $
• B. $10^{-28} $
• C. $10^{18} $
• D. $10^{17} $

Answer 1

Answer: (A)

$Zn (s)+ Cu ^{2+}(a q) \rightleftharpoons Cu (s)+ Zn ^{2+}(a q)$.
$\therefore \quad E^{\circ}=\frac{0.0591}{n} \log _{10} K_{\text {eq }}$
$E^{\circ}=+1.10 V$
because at equilibrium, $E_{\text {cell }}=0$
$(n=$ number of electrons exchanged $=2$ )
$1.10 =\frac{0.0591}{2} \log _{10} K_{ eq } $
$\frac{2.20}{0.0591} =\log _{10} K_{ eq }$
$K_{ eq } =\text { antilog } 37.225$
$K_{ eq } =1.66 \times 10^{-37}$