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The equilibrium constant of the reaction:
$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$;
$E^\circ = 0.46$ V at 298 K is
$Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$;
$E^\circ = 0.46$ V at 298 K is
A
$2.0 \times 10^{10}$
B
$4.0 \times 10^{10}$
C
$4.0 \times 10^{15}$
D
$2.4 \times 10^{10}$
Detailed Solution
At equilibrium $E_{cell} = 0$, so the Nernst equation gives $E^\circ_{cell} = \dfrac{0.0591}{n}\log K_c$ at 298 K.
In the reaction, Cu loses 2 electrons and two $Ag^+$ ions gain them, so $n = 2$.
$\therefore 0.46 = \dfrac{0.0591}{2}\log K_c$
$\log K_c = \dfrac{2 \times 0.46}{0.0591} = 15.57$
$K_c$ = antilog 15.57 $= 3.7 \times 10^{15}$
$K_c \approx 4 \times 10^{15}$
In the reaction, Cu loses 2 electrons and two $Ag^+$ ions gain them, so $n = 2$.
$\therefore 0.46 = \dfrac{0.0591}{2}\log K_c$
$\log K_c = \dfrac{2 \times 0.46}{0.0591} = 15.57$
$K_c$ = antilog 15.57 $= 3.7 \times 10^{15}$
$K_c \approx 4 \times 10^{15}$