How do you finish the reaction: $Zn(s) + CuSO_4(aq) ->$ ?

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Answer 1 · 2017-04-08T00:58:29

$->ZnSO_4(aq)+Cu(s)$

Zinc, being the less noble metal of the two, will donate two electrons to the copper(II)-ions:

$Zn(s)+Cu^(2+)(aq)->Zn^(2+)(aq)+Cu(s)$

How do you finish the reaction: Zn(s) + CuSO_4(aq) ->Zn(s) + CuSO_4(aq) ->?