Zn(s)/Zn2+(aq)//Au+(aq)/Au(s)
Reduced 60%
NO3-
no it can not be reduced.... :)
It cannot be reduced
Answer this question… Fe3+
Fe3 is the least likely to be reduced because it has the highest oxidation state among the listed elements, making it less prone to gaining electrons and being reduced.
Mg2+
Zn2+
Idontremember is right, Na+ was correct for me on Apex
Mg(NO3)2+ZN2+--->ZN2++2NO3-+Mg(s)
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
Zn2+ + 2e- <--> Zno -0.7618 V
It's name is just a zinc ion. Normally its symbol is Zn2+
The white color of Zn2+ complexes is attributed to the fact that Zn2+ lacks partially filled d orbitals for d-d electronic transitions that typically give rise to color in transition metal complexes. As a result, Zn2+ complexes do not absorb visible light in the range that produces color, leading to their white appearance.
Mg(NO3)2+ZN2+--->ZN2++2NO3-+Mg(s)
Mg(NO3)2+ZN2+--->ZN2++2NO3-+Mg(s)