A) \[0.177V\]
B) \[0.087V\]
C) \[0.059V\]
D) \[-0.177V\]
Correct Answer: D
Solution :
\[pH=3,[{{H}^{+}}]={{10}^{-3}}\] \[E=E_{red}^{o}+0.059\log (ion)\] \[E=0+0.059\,\log \,({{10}^{-3}})\] \[E=+0.059\,(-3)=-0.177V\]You need to login to perform this action.
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