A volume of x mL of 5 M NaHCO3 solution was mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to record a cell potential of 235.3 mV, then the value of x = mL (nearest integer). Sn(s) | Sn(OH)62- (0.5 M) | HSnO2- (0.05 M) | OH- | Bi2O3(s) | Bi(s) Consider up to one place of decimal for intermediate calculations [ arrayllGiven: & E Sn(OH)62-|HSnO2- = -0.9 V \ & E Bi2O3|Bi = -0.44 V \ & pKa(H2CO3) = 6.11 \ & 2.303RTF = 0.059 V \ & Antilog (1.29) = 19.5 array ]
Answer
78
Solution
Anode: HSnO2- + 3OH- + H2O Sn(OH)62- + 2e- ( 3)
Cathode: Bi2O3 + 3H2O + 6e- 2Bi + 6OH-
Cell: 3HSnO2- + Bi2O3 + 6H2O + 3OH- 3Sn(OH)62- + 2Bi; n = 6
E cell = -0.44 - (-0.9) = 0.46 V
Nernst equation:
Ecell = E cell - 0.0596 [ Sn(OH)62-]3[HSnO2-]3[OH-]3
= 0.46 - 0.0592 0.50.05 [OH-]
0.2353 = 0.46 - 0.0295 10[OH-]
10[OH-] = 0.22470.0295 = 7.6
1 + pOH = 7.6 pOH = 6.6
pH = 14 - 6.6 = 7.4
Buffer: mmol of HCO3- = 5x, mmol of H2CO3 = 10 2 = 20
pH = pKa + [HCO3-][H2CO3]
7.4 = 6.11 + 5x20
x4 = 1.29 x4 = 19.5
x = 78
Answer: 78