A) \[-0.32V\]
B) \[-1.20V\]
C) \[+1.20V\]
D) \[+0.32V\]
Correct Answer: D
Solution :
\[Z{{n}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Zn;\,\,\,{{E}^{o}}=-0.76V\] \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe;\,\,{{E}^{o}}=-0.44V\] Cell reaction is \[F{{e}^{2+}}+Zn\xrightarrow{{}}Z{{n}^{2+}}+Fe\] \[{{E}_{cell}}={{E}_{cathode}}-{{E}_{anode}}\]c \[=-0.44-(-0.76)\] \[=-0.44+0.76\] \[=0.32V\]You need to login to perform this action.
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