1.
  2Na(s) ------------------> 2Na^+ (aq) +
2e^-Â Â Â Â Â E0Â Â = 2.71v
2H2O(l) +2e^- -------> H2(g) + 2OH^-
(aq)Â Â Â Â Â Â Â E0 = -0.83v
------------------------------------------------------------------------------------
2 Na(s) + 2 H2O(l) --> H2(g) + 2
OH-(aq) + 2 Na+(aq)Â Â Â E0cell
= 1.88v
n = 2
ΔGo   = -nE0cell*F
          =
-2*1.88*96500
          =
-362840J/mole   = -362.84KJ/mole
2.
2Fe(s) --------------> 2Fe^3+(aq) +
6e^-Â Â Â Â Â Â E0 = 0.04v
6H^+ (aq) + 6e^- ------>
3H2(g)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
E0Â Â = 0.00v
------------------------------------------------------------------------------
2 Fe(s) + 6 H+(aq) --> 2 Fe3+(aq) + 3
H2(g)Â Â Â Â E0cell = 0.04v
n = 6
ΔGo   = -nE0cell*F
          =
-6*0.04*96500
            Â
= -23160J/mole
ΔGo   = -RTlnK
-23160Â Â = -8.314*298*2.303logK
-23160 = -5705.84logK
logKÂ Â Â Â Â Â Â Â =
-23160/-5705.84
logKÂ Â Â Â Â Â Â Â = 4.06
KÂ Â Â Â Â Â Â Â Â Â Â Â
= 10^4.06Â Â = 1.15*10^4 >>>>answer
