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