A) cationic part
B) anionic part
C) both in cationic and anionic part
D) there is no zinc in the solution
Correct Answer: B
Solution :
\[Z{{n}^{2+}}+2O{{H}^{-}}\xrightarrow[{}]{{}}Zn{{(OH)}_{2}}\downarrow \]\[Zn{{(OH)}_{2}}(s)+2O{{H}^{-}}(aq)\xrightarrow[{}]{{}}ZnO_{2}^{2-}(aq)\]\[+2{{H}_{2}}O(l)\]You need to login to perform this action.
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