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OpenStudy (anonymous):

What reduction half-reaction happens when zinc reacts with nickel(II) chloride? Zn(s) + NiCl2(aq) ZnCl2(aq) + Ni(s)

OpenStudy (anonymous):

Zn(s) + NiCl2(aq) --> ZnCl2(aq) + Ni(s)

OpenStudy (anonymous):

@Australopithecus @sweetburger

OpenStudy (anonymous):

@Abmon98

OpenStudy (anonymous):

I think it's Ni(s) --> Ni2+(aq) + 2e− but I want to make sure I'm right

OpenStudy (abmon98):

The oxidation state of an uncombined element is zero. \[Zn(s) + NiCl2(aq)\rightarrow ZnCl2(aq) + Ni(s)\] Zn(s)-->Zn+2(aq)+2e- Ni(+2)+2e--->Ni

OpenStudy (anonymous):

oh so it's Zn(s) --> Zn2+(aq) + 2e−

OpenStudy (abmon98):

yes thats right

OpenStudy (anonymous):

Ni(+2)+2e--->Ni

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