Question

19. Calculate the standard free energy change for the following reaction: 4AH + O218) + 4H (39) - 4Ag" (g) + 2H₂0m. 20. Calculate the number of grams of aluminum produced in 2.00 hour by the electrolysis of molten AlCl if the electrical current is 15.0 A. (IF = 96.500 C/mole) 21. Calculate the standard free-energy change for the following reaction at 25 °C: Cup) + Znis) --> Cu + Zn (aq) (F=96500 J/Vmol)

Answer 1

19. Gibbs Free Energy change can be calculated from the enthalpy (∆H) and entropy changes (∆S) as follows. (NOTE: Values for enthalpies and entropies are obtained from standard thermodynamic data at 298 K).

(19 Standard tree energy change (AGⓇ) AGⓇ - AH-TAS - © Change in enthalpy (AH) TH={products - Ehreactants ZH products = 4x4H f (Agt) + 2 x 4H f (H₂O) = (4x 105.58) + (2 X-285.83) = 422.32 + (-571-66) = -149.34 kJ mol-1 ZFgeactants = 4x AHF + oHere) + 4 x AHFCH+) = (4x0) + (0) + 4*0) = 0+0+0 = 0 kJ molt → AH =(-149.34_0) kJ mot' = -149.54 kJ mot- Since the temperature is not specified, assume it to be 298 K.

biu. Change in entropy (as) - AS = ES products - ES reactants Esproducto = 4 XSagt + 294,0 - (4 X7268) + (2x69.91) = 290.72 + 139.82 = 450.54 J K mot IS reactants = AXSAg & So, + 4x 54 = (4x42.55) + . 205.14 + (4x0) = 170.2 + 205.14 to the = 375.34 Jkt mot ogoo al

AS = (430.54 - 375.34) J K mol- = 55.2 J K - mot - J K molto Substituting the v in the above equ We have tuting the values of entropy and enthalpy above equation for Gibbs free energy change, 46 - (-149.34X103 J mol-1)_ (298 K x 55.2 JK 'mol-') ch bromsen = -165789.5 J mol- o AG° = -165.7895 kJ mol'] in Since 4G for the reaction has a negative value, the reaction is spontaneous,

20. Amount of Al produced during electrolysis can be calculated as follows.

(20) Molten AlCl is used as an electrolyte. It would dissociate to give, AlCl3 - Al3+ + 301 1 molecule - lion of AlCl3 of Alat As we can see from the stoichiometry , 1 molecule of AlCl3 gives lien of Al3t. Also stope Al3+ + 30 Ad 1 mol 3 I 1 molof X - - 4 Al3+ Al. This implies that we need to provide 3 of current per mole deposition of Al. Also note that 1 mole of Al weighs 27 g.

» 3F of current required to deposit 27g of Aluminium. This also means 289500 C of change is required to deposit 279 of Al. (1F = 96500c) Given Current = 15.0 A Time =2.00 hours - 72005. - Change (Q) = Current & Time - 15 A X 7200 5 = 108000 Now, 289500C 108000 C charge of Etettent deposite 279 of Al. of charge will deposit x' of Al >> X = 279 x 108000 C 289500C = 10.072 9. Hence 10.072 g of Al will be produced in 2 hours,

21. Standard free energy change can be calculated as follows.

21 Standard free energy change (46°) AG = nF E cell where n= number of elections involved. in half reactions. F-1 Faraday = 96500 C Fem - Standard EMF of the cell.

cu + Zn (6) C + Zn (aq) (aq (Loss of elections) As we can see, Oxidation half reaction Zn 6 n tag) the Reduction half reaction Cuidas + 20 - Cu(s) (Grain of elections) Oxidation half reaction occurs at the anode and the reduction half reactions occur at the cathode. Ecell - Ecathode -Eanode = Ecu" /cu - E Zn/Zn = 0.34 V - (-0.76V) NOTE Data obtained from standard reduction potential at 298k. = 1.1 V Also n=2 - 1F = 96500c Substituting the values obtained in AGiº expression We have, I NOTE AG -2x 96500c X 1.1 V T C = 15V molt --212300 J mol- AG =-212.3 kJ mol-1 As the agiº value ºs negative, the reaction is feasible at 25€.