What is the Ph of a 3.9* 10^-8 M OH- solution
1 answer:
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Answer:
a) galvanic cell
b)electrolytic cell
c) i) K=6.27x10'34
ΔG°=198790 J
ii) K=3.58x10'-34
ΔG°= 191070 J
d) E°=0.278 v
ΔG°= -26827 J
Explanation:
a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".
The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.
b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.
c) i) look image attached
ii) k = look image attached
ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)
ΔG° =-191070
d) E°= 0.0592 v/n x lg K
E°= 0.0592V / 1 X log 5.0X10'4
E°= 0.278 v
ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v
ΔG° = -26827 J
Answer:
The pH of the solution is 5.31.
Explanation:
Let " is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.
Initial C 0 0
Equilibrium c(1- ) c
c
Dissociation constant =
In this case weak acids is very small so, (1-
) is taken as 1.
From the given the concentration = 0.050 M
Substitute the given value.
Therefore, The pH of the solution is 5.31.