Use tabulated electrode potentials to calculate δg∘ for the reaction. 2na(s)+2h2o(l)→h2(g)+2oh−(aq)+2na+( aq)

Na --> Na+ + e- E = +2.71 2H2O + 2 e- --> H2 + 2 OH- E = -0.83 Delta E = 2.71 - 0.83 = 1.88 Volts You know the reaction is spontaneous because Ecell is positive. Delta Go = - n F Delta E Delta Go = - 2mol (96485 J/volt mol) (1.88 V) Delta go = -3.62X10^5 J/mol = -326 kJ/mol Yes, it is VERY spontaneous.