a)Calculate the potential for the reaction below at 25ºC under
the following conditions:
[MnO?¯ ]=0.0200M
[Ni+2 ]=2.50M
pH = 2.000
2 MnO?¯ (aq) + 8 H+ (aq) + 3 Ni (s) ? 2 MnO? (s) + 4 H?O
(l) + 3 Ni+2 (aq)
Eocell = Eocathode - Eo anode
= 1.68 + 0.26
= 1.94 V
Q = [Ni+2]^3 / [MnO4-]^2 [H+]^8
Q = (2.50)^3 / (0.02)^2 (10^-2)^8
Q = 3.91 x 10^20
Ecell = Eo cell - 0.05916 / n log Q
= 1.94 - 0.05916 / 6 * log (3.91 x 10^20)
= 1.74 V
Ecell = 1.74 V
a)Calculate the potential for the reaction below at 25ºC under the following conditions: [MnO?¯ ]=0.0200M [Ni+2...
Calculate the potential for the reaction below at 25ºC under the following conditions: [Fe+3 ]=0.0200M [Sn+2 ]=0.0200M [Fe+2 ]=1.50M [Sn+4 ]=1.50M 2 Fe+3(aq) + Sn+2 (aq) → 2 Fe+2 (aq) + Sn+4 (aq)
Calculate the potential for the reaction below at 25ºC under the following conditions: [Fe+3 ]=0.0200M [Sn+2 ]=0.0200M [Fe+2 ]=1.50M [Sn+4 ]=1.50M 2 Fe+3(aq) + Sn+2 (aq) → 2 Fe+2 (aq) + Sn+4 (aq) E = Answer V
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