Consider the following electrochemical cell at standard condition. Au(s) | QH2, Q | NH4X(0.01 M) || Ag+(1 M) | Ag(s) Ecell = +0.4 V The couple QH2/Q represents quinhydrone electrode, the half cell reaction is given below: [figure] [Given : E0 Ag+/Ag = +0.8 V and 2.303RTF = 0.06 V] The pKb value of the ammonium halide salt (NH4X) used here is . (nearest integer)
Answer
6
Solution
Anode (left, oxidation): QH2 Q + 2H+ + 2e-
Cathode (right, reduction): 2Ag+ + 2e- 2Ag
Reduction potential of the quinhydrone electrode:
EQ/QH2 = E Q/QH2 - 0.062 1[ H+]2
= 0.7 - 0.06 pH
Ecell = E Ag+/Ag - EQ/QH2 (with [ Ag+] = 1 M, E Ag+/Ag = 0.8 V)
0.4 = 0.8 - (0.7 - 0.06 pH)
0.06 pH = 0.3
pH = 5
NH4X is a salt of a weak base and a strong acid (hydrolysis):
pH = 7 - 12pKb - 12 C
5 = 7 - 12pKb - 12 (10-2)
5 = 8 - 12pKb
pKb = 6
Answer: 6