Question

3. When C12(g) is added to aqueous K4[Fe(CN).), [Fe(CN)]*- is oxidized to [Fe(CN).]>- and K:[Fe(CN)] precipitates, as shown below. 2K.[Fe(CN),](aq) +Cl2(g) = 2Ks[Fe(CN)2](3) + 2Cl(aq) a. Write the net ionic equation for the reaction and calculate its equilibrium constant. (10 pts) At 25°C [Fe(CN).J” (aq) + € → [Fe(CN).]" (aq) E = 0.356 V Cl2(g) + 2e → 2Cl(aq) E* = 1.360 V Kup of Kj[Fe(CN).] = 106.5 Answer:

Answer 1

3. (a) Given, 2 Ky [Fe (CN). I aq + c g) = 2K [Fe(CN) ICS) + kacaq? ak@pt 2 Cfe Cen), je coq) + 499) * -zeretetet 2 kg [fe CCN) (s) + 2 k caps Remering the common species both side, we get the net lonic equation. 6 Kecogs + 2 (FeCen), J4@caq) + 4949) = 2 Kg CFe CCN), ] (s) + 20° (05) + 20 che copy Now, - 2x [Fe(CN) 73 Cag) = 0.356 v [Fe(CN) 740 (aq) 6 = 201® (aq) E = 1.360v " (g) + 200 2(Fe (CN), 3*97) + (8) aC Fe Cam), J* Caps + 20%) Eiell = Ecathode - Eanode = Ereduction - Eoxidation Ecelu = Eran = 1.004v = 1- 360 V-0.356 v We know that DG = -RT Inkog = -nff - linkeq= NFERAN & log keq nFETAN 2.303 RT At T = 25°C RT 2.303 RT F = 0.0592 V here, n=2 So 2 X 1.004 V log keq = = 33.92 33.92 2) keq = 10 = 8.30x1033 0.0592 V Now, the net ionic equation of desired reaction. ks [Fe(CN) J () = 3 Koppt [fe Cen), 3*Cap Ksp = 106.5 - Multiplying equation & with 2 and sutittract from , we get Net bonic equation GK@caq) + 2 Cfecemno omlag) + (1999 -> 2 [feCan). Jo® g) +24%) Keg 8.30 x1033 = 7.31x1029 Ks (106.58 The equilibrium constant for net ionic equation (ky) = 7.31x10 29 Using nandito addition and substration rule for equilibrium Constant keq

(6) Giver, concentration of KON = 1.00M , volume of KCN 30.0mL amount welas of KCN added 4.00 Đo KCN 800 850 kcN X = 80.0 mmol KCN 1.00 ml KON Concentration of Fell = 0.020om, volume of felly = 20.0 ml * 0.0200 mmol Fede amount rooted of fell added = 20.0 ml fell, 1.00 ml Fella = 0.400 mmol Fell We know that ALE 6 KCN Cap) + Fech cas) = K[Fe(CN),] Caps + 2 K4 Cay) Ionic equation GCNO Cap + Fe 2009) (fe CCN) J 40 cap) As ko and ae are not particitating in the above reaction, the concentration of Ko and ce are as follows = 0.800 M [ko] = 80.0 mohol Total volume of Solution = 80.0+ 20.0 = 100.0ml 100.0 ml 2 X 0.400 mmol [co] = -0.00909 1909 note! Pe cl,cop) -Fe 284 28 10o.oml 1 mol 2 met Here, limiting reagent is fecl . So, the formation of CFe (CN) to cap is based on Fello 0.400 monol 100.0ml = 0.00400 M [fe CCN) 40 We know that If for EfeccN), = 1.00 1035 at T = 25°c As this value is too large which mean diesociation of (fe COND] 4 to Fe20 and one is negligitable. Hence, equilibrium concentrations of Ko cho and (FeCCM) 740 ces follows K. 0.800 M ce: 0.00 800 M The CCN) 4+ 0.00400M are

If 0.800 0.00400 0.00800 0.100 ml 10ooml = 0.00100 1 - 6x - 2x +2x 0.00400- 2x 0.0010o-2 0100502 0.00800 +2x 0.100 monol a cg) re added the part b, the below greaction takes place 6K@cag) +2Cfe Cendo J4C09) + 499) = alfecon)g7g) +24160p Initial concentration (m) change in concentration Equilibrium concentration 0.800 - 6x (M) Facenda cuen? kep [ KB) & { [ Fe CCN) 79072 [leg) (0.00800+ 2x72 7.31x1029 (0.800-6x) (0.00400-2x)² (0.00100-x) As the value of keg is too large, the reaction will go to the for forward direction. Here limiting reagent is Ugg) reacts with 2 mol of Fe(CN) 0.0010000) M 4,9 sneach with B8-9X0-00100M of [FOCIN) 70 2 [Fe(CN) 73() +200 CU Initially concentration sem & Change in concentration +62 Equillonum concentrat 0.794+62 egottee 1 mol 4,9 qe - 6.00200M Crecente + 2[Fe(CN), Acap) + 6ko 0.00800 +0.00200 0.800 +0.00600 = 0.0100M 0 0.00400- 0.0020 -0.00200 0.79% +2x 0.08100-2x 0.00200+2x Ck@j®(CFe Cen), Jde) cu,92 x 1.37x10.30 ket Cooja (0.794 +6x) (0.00200 +2 x (x) 7.31x1029 (6.0100-2x32 x < 0.794, 0.00200, 0.0160 1637 41030 (0.794) (0.002002 1.37x10 30x (0.01008 (0.0100)? 10-7949% (0.0020012 a x = 1,36x1020 m

The equilibrium concentration of Cl2(g) is 1.36x10^-20 M

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