Zn(s)/Zn2+(aq)//Au+(aq)/Au(s)
Reduced 60%
NO3-
no it can not be reduced.... :)
No it cannot be reduced
Zn2+
Mg2+
Zn2+
Idontremember is right, Na+ was correct for me on Apex
Answer this question… Fe3+
For emample, if you are dealing with Copper and Zinc, you first take the half equations Cu2+ + 2e- --> Cu +0.34 Zn2+ + 2e- ---> Zn -0.76 then, you decide which one is reduced and which one is oxidized, the one with the more positive voltage is reduced, so in this case the copper is reduced. the overall reaction changes to Zn- ---> Zn2+ + 2e +0.76 Cu2+ + 2e- --> Cu +0.34 ---- Zn + Ni2+ ---> Ni + Zn2+ + 0.5 you add the voltages and get 0.5 as an experimental voltage
Cu+ is isoelectronic with Zn2+
Mg(NO3)2+ZN2+--->ZN2++2NO3-+Mg(s)
Zn2+ + 2e- <--> Zno -0.7618 V
Zinc forms Zn2+
It's name is just a zinc ion. Normally its symbol is Zn2+
Mg(NO3)2+ZN2+--->ZN2++2NO3-+Mg(s)