Question

Tabje 1: Standard Reduction Potential at 25°C Jadual 1: Potensi Penurunan Standard pada 25 C Halt-Reaction +2K7 +207 +L82 +1.77 +1.70 2 (ng EB8160 11- Ogt4 211 2r 0te l4 Co HOng2H n) 2r 2114 PO4+41lng SO tagt2 Ce ng MaO, tay) + sli tay+ Au te) C14et2 C10 tangt 1411 t Mot)4s a1 tng 2 04e1+411 tety 4t Brals2 2Br tg NO. tay) +411 ()+ Determine whether the following reaction would proceed spontaneously PhS0A 2H0 as written at 298 K Refer to Table 1 for further details ce a +1.51 +1.50 +1.16 Maert+ 411-0 Antsi Tentukan sama ada tindak balas berikut akan berterusan secara spontan -20 tg)711 Me ty) 140 1.11 seperti yang deuls pada 298 K Rujuk pade Jadual 1 untuk maklumat 1.23 lanjut +1.23 +107 96 NOKg)2114 Cola)-Fe(a) Co(aFels) +092 2Hpth Ages) Hgtay 2e Ag'tny +0.9 Cols)+Fe(ak) Ce(ak)eFels) +0 77 +0.68 Oe21 says2r- Mat) fy + 210 e Ma04at +40 (y 14a+2 21 tah Oge)+2114 de +059 +053 Given that Ce)-0.15 M and Fel")-0.68 M +010 +L14 Di ber -0.22 Apt t+e SO tag)4H a &Ctng Agts)+(7 0.20 S04gt 214 Co"-0 15 M and Fe")0 68 M 0,15 (30 marksimarkat) 0.13 Se tun 0.13 -014 S (ongt2 Ni t 2 Co'g)-2e hsDAn2e Cdtag)+ 2e Cak Ie ta+e Fetst Cr' tan C'a Za t)+2e 2314) 2 I14e) 20HI Mu teg+ 2 Aag+e Be ta 2e Mg de Mgia Na'tag)+ Ca tan+e Sr'a 2 Snt 025 Nits Ph Sta -0.11 -0-40 -144 0.74 -a76 Zai Matsl -118 Ash -1.85 -2.37 -2.7 -2 87 Nat Catx Sts -2.90 -293 Ba int K ag Li togi Lats Bata Ktst -105 he 4hdlsr the mled sta d Mste due qilas BureaN

Answer 1

Solution:

For a reaction to proceed spontaneously, the change in free energy (ΔG) must be negative.

The value of ΔG related with emf of the cell as,

ΔG = - nFE

Thus, a negative value of ΔG means, ΔE must be positive.

For the given cell:

Co (s) + Fe2+ (aq) = Co2+ (aq) + Fe (s)

The cell consists of two half cells,

Co (s) = Co2+(aq) + 2 e- ( oxidation)

Fe2+(aq) + 2e- = Fe (s) ( reduction)

Net reaction,

Fe2+ (aq) + Co(s) = Fe (s) + Co2+ (aq)

The standard reduction potentials (E°) are given in table as,

Co3+ / Co = - 0.28 V ( E°R)

Fe2+/Fe = -0.44 V ( E°L)

E°cell = E°R - E°L = - 0.28 - (- 0.44) = 0.16 V

According to Nernst equation:

E = E°cell - (0.0592 / n ) log [Co2+/ Fe2+]

E = 0.16 - (0.0592 /2) x log (0.15/0.68)

E = 0.16 V - 0.0296 x log 0.22

E = 0.16 V - 0.0296 x (- 0.658)

E = 0.16 V - 0.019 = 0.179 V

Since, E = +ve,

Hence, ΔG = -ve

Therefore, reaction is spontaneous.