Question

The table below gives conductivity data at infinite dilution measured at 25 °C. Electrolyte Aº/mS mmol.' KOOCH 12.81 HCI 42.60 KC 14.99 (a) Why can the limiting molar conductivity of methanoic (formic) acid not be measured directly? (b) Calculate the limiting molar conductivity of methanoic acid. The molar conductivity of 0.020 M HCOOH(aq) is 3.83 mS mmol"!. What is the pk, value for methanoic acid? The following questions are about electrochemistry, Explain how the electrochemical series can be used to determine if a redox reaction is spontaneous. Consider the cell shown below: Fe(s)| Fes(aq) || Ag (aq) Ag(s) (1) Write both the half-cell and the full cell reactions, (ii) Calculate the cell potential at 298 K if the cell is no longer under standard conditions, but has the following concentrations: Fe(s) Fe (aq, 0.20 mol dm) || Ag (aq, 0.050 mol dm) Ag(s) Fe (aq) + 2e - Fe(s) Ag (aq) + e - Ag(s) 0.44 V +0.80 V (iii) Calculate AG for the concentration cell in part (ii).

Answer 1

(a) Formic acid do not dissociate completely and it's variation does not increases linearly . Hence we can not extrapolate the graph of Molar conductivity vs concentration .

im. HCOOH = K DOCH Caz. = 12.81 +42-60-14.99 = 40-42 ms. m mol (Answer) 2= nm, Hoon 3-83 ms m Imol = 0.094755. "m, neoon 40.42 ms m/mol неон е н® + HP Clock - -0.020x (0.044 ((1-2) CR ca ((-) 1-0.094755 Ka = 1.9837X10 blac-logka = - log (1.9837x154) = 3.70 ( Answer) (d) A species which has standard reduction potential, would gets reduced Eccle should be higher bositive. Age-n fecell Ago would be negative and reachön would be spontaneous. © is anode Rxn fers) fe2t+ae cathode Rxn lap) t ë Agrs) full cell reaction: Fe(s) + 2 Ag@ cap) → fet cape) + 2 Ag(s). (0 cele = +0-80 V-1-0-449 = 1-24V Ecele cele - RT en fet Ecole = 1.24 - 8.314x 298 en (0030)2) = 1.24-0.0626 = 1.1837 : (Ag+] 2 2x96485 = 1.18 V.(Anwer) (1) DGz-nf Ecer = -296485 x 118 = -228426-01J mol AG = -228.4 kJ/mol. (Aonwer)