A) \[0.94V\]
B) \[0.76V\]
C) \[0.40V\]
D) \[0.20V\]
Correct Answer: D
Solution :
\[E=E{}^\circ -0.06\log \frac{[{{H}^{+}}]\left[ C{{l}^{-}} \right]}{{{[{{H}_{2}}]}^{1/2}}}\] \[0.92=E{}^\circ -0.06\log \frac{{{10}^{-6}}{{.10}^{-6}}}{{{1}^{1/2}}}\] \[0.92=E{}^\circ -0.06\log {{10}^{-12}}\] \[=E{}^\circ -0.06\times 12\] \[E{}^\circ =0.92-0.06\times 12\] \[E{}^\circ =0.92-0.72\] \[\frac{E_{Ag}^{o}}{AgCl}=0.20.\]You need to login to perform this action.
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