A) \[Cu+F{{e}^{2+}}\xrightarrow{{}}C{{u}^{2+}}+Fe\]
B) \[2Cu+F{{e}^{2+}}\xrightarrow{{}}2C{{u}^{+}}+Fe\]
C) \[Fe+C{{u}^{2+}}\xrightarrow{{}}F{{e}^{2+}}+Cu\]
D) none of the above
Correct Answer: C
Solution :
From the cell representation, it is clear that Fe makes the anode and Cu makes the cathode of cell. Hence, Fe will undergo oxidation while Cu reduction. Therefore, half-cell reactions will be \[Fe\xrightarrow{{}}F{{e}^{2+}}+2{{e}^{-}}\] ?(i) and \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] ?(ii) Hence, the cell reaction will be \[Fe+C{{u}^{2+}}\xrightarrow{{}}F{{e}^{2+}}+Cu\]
You need to login to perform this action.
You will be redirected in
3 sec
