Chemistry redox reaction

Question

Calculate the E^o cell  for the redox reaction shown below in volts:

2Li(s)+Mg+2(aq)→2Li+1(aq)+Mg(s)2

Eocell=Eocathode+Eoanode

Answer 1

2Li(s) + Mg²⁺(aq) ==> 2Li⁺(aq) + Mg(s)

From a table of standard reduction potentials, I find the following values:

Li⁺ + e- ==> Li Eº = -3.05 V

Mg²⁺ + 2e- ==> -2.37

Mg will be reduced and so will be the cathode

Li will be oxidized and so will be the anode

Eº_cell = Eº_cathode - Eº_anode = -2.37 - (-3.05) = -2.37 + 3.05

Eº_cell = 0.68 V

Answer 2

In this reaction (as written) Li is undergoing oxidation (anodic reaction, it is always oxidation taking place at the anode), and Mg²⁺ is undergoing reduction (cathodic reaction, it is always reduction taking place at the cathode).

the two std. red. potentials are: Mg_(aq)²⁺ + 2e^- ====> Mg(s). E^o = - 2.38 V

Li_(aq)⁺_. +. e^- ====> Li _(s) E⁰ = -3.04 V

Std. cell Potential. (E^o_cell) = E^o_cathode - E^o_anode. =. -2.28 - (-3.04) = 0.66 V.