6. (22 pts) A galvanic cell is set up to measure the acid dissociation constant

Question

6. (22 pts) A galvanic cell is set up to measure the acid dissociation constant (K) for the weak acid HA. The cathode consists of a (Ag./Ag) half-cell ('+0.800 v) in which I Ag.|e 1 00 M. The anode consists of a Pt electrode dipping into the solution and H2(g) is bubbled into this solution at a pressure of 1.0 atm. The anode half-cell (H30+/H2E-0000 V) solution contains 0.10 M HA(aq) and 0.050 M A (a) its conjugate base. The potential for this cell (Ae) is found to be 1.030 V at 25 C. (Note: HA and A do not participate in any redox reactions in the anode half-cell.) a) Write down the two half cell reactions and the overall net cell reaction. b) Determine the H30 (ag) concentration in the anode half.cell c) Determine the acid dissociation constant

Explanation / Answer

a) cathode half-cell

Ag+  + e- ----> Ag............(x2)

anode half cell

H2(g) ----> 2H+ + 2e-

overall reaction

2Ag+ + H2 + 2e- ----> 2Ag + 2H+ + 2e-

b) Eocell = Ered +Eoxd

  = 0.8 + (-0)

= 0.8V

delEcell= E0red + Eooxd

1.03= 0.8 + Eoxd

Eoxd= 0.23

now, Eoxd= E0 - [(0.0591/n) * log (1/[H+])}

= 0 - {0,0591/2 * log (1/[H+])}

0.23 = -0.0295 * log (1/[H+])

log (1/[H+])= -7.79

log [H+] = 7.79

[H+] = 6 * 107M

c) Ka= {[conjugate base]*[H+]/[acid]}

= 0.05*6*107/0.1

Ka=3 * 107

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