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29 protons and 27 electrons.
29 protons and 27electrons are present in Cu2+ ion.
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
The electron configuration of Cu2+ is [Ar]3d94s0.
.727/g*M, slope is molar absorbtivity
Cu2+
Answer this question… Fe3+
Cu2+
Zn2+
something about "native" should be ur answer..apex
Cu2+ + I- --> Cu2I The compound created is Copper(I) Iodide
"All monosaccharides and most disaccharides can be oxidized. When the cyclic structure opens, the algehyde (-COOH) group is availabel for oxidation. Benedict's reagent contains Cu2+ ion that is reduced. Therefore, all the sugars that react with Benedict's reagent are called reducing sugars. When oxidation of sugar occurs, the Cu2+ is reducted to Cu+, which forms a red precipitate of cuprous oxide, CuO(s). The precipitates colors varies from green to fold to red depending on the concentration of the reducing sugar." Caralyst Pearson Custom Publishing 2011 p.65
Cu(s) | Cu2+(aq) Au+(aq) | Au(s)
Copper two
Cu + Mg2 --------> Cu2 + Mg Cu --------------> Cu2 + 2e Mg2 + 2e --------> Mg Cu --------------> Cu2 + 2e (E = +0.35) Mg2 + 2e --------> Mg (E = -2.36V) +0.35 + (-2.36) = -2.01V --------------------------------------… Mg + Cu2 --------> Mg2 + Cu Mg --------------> Mg2 + 2e Cu2 + 2e --------> Cu Mg --------------> Mg2 + 2e (E = +2.36V) Cu2 + 2e --------> Mg (E = -0.35V) +2.36 + (-0.35) = +2.01V
The octet rule does not apply to transition metals.
The overall voltage for a redox reaction with the half reactions Mg s -- Mg2 plus plus 2e- and Cu2 plus -- Cu is 76 V.