| \[ZnZ{{n}^{2+}}_{(aq)}+2{{e}^{-}}\] \[{{E}^{o}}=+0.76\,V\] |
| \[FeF{{e}^{2+}}_{(aq)}+2{{e}^{-}}\] \[{{E}^{o}}=+0.44\,V\] |
| The emf for the cell reaction |
| \[F{{e}^{2+}}+ZnZ{{n}^{2+}}+Fe\]will be |
A) \[-1.20\,V\]
B) \[0.32\,V\]
C) \[-0.32\,V\]
D) \[-1.2\,V\]
Correct Answer: B
Solution :
\[Z{{n}_{(s)}}Z{{n}^{2+}}_{(aq)}+2{{e}^{-}},\]\[{{E}^{o}}=0.76\,V\] (Anode) \[F{{e}_{(s)}}F{{e}^{2+}}_{(aq)}+2{{e}^{-}},\] \[{{E}^{o}}=+0.44\,V\] (Cathode) \[{{E}^{o}}_{cell}={{E}^{o}}_{anode}-{{E}^{o}}_{cathode}\] (In oxidised electrode potential) \[=0.76-0.44=0.32\text{ }V\]
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