How do you finish the reaction: $Z n (s) + C u S O_{4} (a q) \to$ $Zn(s) + CuSO_4(aq) ->$ ?

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Answer 1 · 2017-04-08T00:58:29

$\to Z n S O_{4} (a q) + C u (s)$ $->ZnSO_4(aq)+Cu(s)$

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

$Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$ $Zn(s)+Cu^(2+)(aq)->Zn^(2+)(aq)+Cu(s)$

How do you finish the reaction: Zn(s)+CuSO4(aq)→Zn(s) + CuSO_4(aq) ->?