Solution to PS 4 #10
a) DpH=7.5-6.7=0.08
By definition, Dp=DG/nF (n=1 in this example, and Eqn. 14.5 has also assumed n=1)
Inserting DG from Eqn 14.5 gives; Dp= DY-(2.303RT)DpH
and at 25 C, 2.303RT/F= 0.059 v so:
Dp=DY - (0.059
V)DpH=-0.18 V - (0.059 V)(0.08)= -0.23 V
b) Dpchem=[(0.059 V)(0.8/-0.23
V ]x 100%= 21%
Dpelec= [-0.18V/-0.23 V] x 100%=
78%
c) DG =nFDp= (1)
(96.48 kJ V-1 mol-1(-0.23 V) = 22 kJ mol-1
Return to Problem Set #4