A) \[5\times {{10}^{6}}\]
B) \[1.4\times {{10}^{12}}\]
C) \[7.4\times {{10}^{12}}\]
D) \[1.2\times {{10}^{6}}\]
Correct Answer: D
Solution :
Cell reaction\[2Cu_{(aq)}^{+}\xrightarrow[{}]{{}}C{{u}_{(s)}}+Cu_{(aq)}^{2+}\] According to Nernst equation \[E=E{}^\circ -\frac{2.030RT}{nF}\log \left[ \frac{M}{{{M}^{n+}}} \right]\] \[E=\]Electrode potential of the metal \[E{}^\circ =\]Standard electrode potential \[R=\]gas constant (8.314 joules per mol per degree absolute) \[T=\]Temperature on Kelvin scale \[n=\]Number of electrons involved in the half cell reaction \[F=\]One Faraday (96500 coulombs) at 298 K, 2.303 RT/F = 0.059 \[\therefore \] \[E=E{}^\circ -\frac{0.059}{n}\log \left[ \frac{M}{{{M}^{n+}}} \right]\] \[=E{}^\circ -\frac{0.059}{n}\log \left[ \frac{Products}{Reactants} \right]\] \[E=0\]at equilibrium \[0=E{}^\circ -\frac{0.059}{n}\log {{K}_{eq}}\] Or \[\log {{K}_{eq}}=\frac{nE{}^\circ }{0.059}\] Or \[\log {{K}_{eq}}=\frac{1\times 0.36}{0.059}=6.09\] \[{{K}_{eq}}=1.2\times {{10}^{6}}\]
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