A) \[1.47\,V\]
B) \[1.77V\]
C) \[1.87V\]
D) \[1.57V\]
Correct Answer: D
Solution :
The half reaction are \[F{{e}_{(s)}}\xrightarrow{{}}F{{e}^{2+}}_{(aq)}+2{{e}^{-}}\times 2\] \[{{O}_{2(g)}}+4{{H}^{+}}+4{{e}^{-}}\xrightarrow{{}}2{{H}_{2}}O\] \[2F{{e}_{(s)}}+{{O}_{2(g)}}+4{{H}^{+}}\xrightarrow{{}}2F{{e}^{2+}}_{(aq)}+2{{H}_{2}}{{O}_{(l)}}\]\[E={{E}^{o}}-\frac{0.059}{4}\log \frac{{{\left( {{10}^{-3}} \right)}^{2}}}{{{\left( {{10}^{-3}} \right)}^{4}}\,(0.1)}=1.57\,V\]You need to login to perform this action.
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