A) increased by 0.059 V
B) decreased by 0.059 V
C) increased by 0.0295 V
D) decreased by 0.0295 V
Correct Answer: B
Solution :
\[F{{e}^{+3}}+{{e}^{-}}\xrightarrow{\,}\,F{{e}^{+2}}\] \[E={{E}^{0}}-\frac{0.059}{1}\log \,\frac{F{{e}^{+2}}}{F{{e}^{+3}}}\]\[{{E}^{0}}-0.059\]You need to login to perform this action.
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