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Half-cell potentials and equilibrium
- Thread starter doctor0404
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1)Ecell (cell potential) = E cathode + E anode
here cathode is where reduction is taking place
Cr^3+ + 3 e- --> Cr (E0 = -0.74 V)
anod where oxidation will take place
Zn - --> Zn^2+ + 2 e(E0 = +0.76 V)
Ecell = -0.74 + 0.76 = +0.02V
spontaneous reaction
2) 2NO2 <===> N2O4
Kc = [N2O4] / [NO2]^2 ----(1)
[NO2]= 0.5M
Kc is also given
just plug in the values in the eq 1
here cathode is where reduction is taking place
Cr^3+ + 3 e- --> Cr (E0 = -0.74 V)
anod where oxidation will take place
Zn - --> Zn^2+ + 2 e(E0 = +0.76 V)
Ecell = -0.74 + 0.76 = +0.02V
spontaneous reaction
2) 2NO2 <===> N2O4
Kc = [N2O4] / [NO2]^2 ----(1)
[NO2]= 0.5M
Kc is also given
just plug in the values in the eq 1
