A) -0.32V
B) -1.20V
C) + 1.20V
D) 4-0.32 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 \[F{{e}^{2+}}\,\,+\,\,Zn\xrightarrow{{}}Z{{n}^{2+}}\,+\,Fe\] \[{{E}_{cell}}={{E}_{cathode}}-{{E}_{anode}}\] \[=-0.44-\text{(}-0.76\text{)}\] \[=-0.44+0.76\] \[=0.32\text{ }V\]You need to login to perform this action.
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