A) \[Z{{n}^{2+}}+2{{e}^{-}}\to Zn(s)\]
B) \[Zn(s)\to Z{{n}^{2+}}+2{{e}^{-}}\]
C) \[M{{n}^{2+}}+2{{e}^{-}}\to Mn(s)\]
D) \[Mn(s)\to M{{n}^{+}}+{{e}^{-}}+1.5V\]
Correct Answer: B
Solution :
At anode: \[Z{{n}_{(s)}}\to Z{{n}^{2+}}+2{{e}^{-}}\].You need to login to perform this action.
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