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