A) - 22.57
B) \[1.6\,\,\times \,\,1{{0}^{-}}^{10}\]
C) \[1.6\,\,\times \,\,1{{0}^{10}}\]
D) \[1.6\,\,\times \,\,1{{0}^{9}}\]
Correct Answer: C
Solution :
Half cell reactions are (i) \[\operatorname{AgC}l \xrightarrow{{}}\,A{{g}^{+}}\left( aq \right) + C{{l}^{-}}\,\left( aq \right)\,\,+ E{}^\circ = 0.22\,V\] (ii) \[A{{g}^{+}}\,+e\,\,\xrightarrow{{}}\,\,Ag,\,\,E{}^\circ =\,\,0.80\,V\] Subtracting Eqs. (i) and (ii) \[\operatorname{AgC}l \xrightarrow{{}}\,\,A{{g}^{+}}+ C{{l}^{-}}\] \[E_{cell}^{{}^\circ } = 0.22 - 0.80 = - 0.58 V\] We know that \[\Delta G = -nFE{}^\circ \] \[-\,RT\,\,In\,k\,\,=\,\,-nFE{}^\circ \] In \[k=-\frac{nFE{}^\circ }{RT}\] \[=-\frac{2\times 96500\times -\,0.58}{8.314\,\,\times \,\,300}\] \[k=-\,22.57\] \[k=\,\,Antilog\,\,\left( \frac{-22.57}{2.303} \right)\,\,=\,\,1.6\times 1{{0}^{-10}}\]You need to login to perform this action.
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