Fe3+(aq) + 6CN-(aq) Fe(CN)63-(aq)
in what sense can the CN- be described as a base and Fe3+ as an acid in this reaction?
Only in Lewis acid --> electron pair acceptor
and the Lewis base --> electron pair donor
in this case
CN is donating 3 electrons (the base)
Fe+3 is accepting 3 electrons (the acid)
Fe3+(aq) + 6CN-(aq) Fe(CN)63-(aq) in what sense can the CN- be described as a base and...
Identify the Lewisacid that acts as a reactant in the following reaction Fe(H2O)63+(aq) + 6CN-(aq) →Fe(CN)63-(aq) + 6H2O(l). Fe(H2O)63+ H2O Fe3+ CN- is it the Fe3+ or the CN-,,,kinda of confused...
The formation constant, Kf, of Fe(CN)63-(aq), Fe3+(aq) + 6 CN-(aq) ⇌ Fe(CN)63-(aq), is 1.0 x 1042. What are the equilibrium concentrations of Fe3+(aq), CN-(aq), and Fe(CN)63-(aq) if we add 0.113 mol of Fe(NO3)3 to 1.000 L of a 0.965 M aqueous solution of NaCN? Assume the volume remains fixed at 1.000 L.
Identify the Lewis acid in the following reaction: Fe3+(aq)+6H2O(aq)?Fe(H2O)63+(aq)
Calculate the overall formation constant for Fe(CN) given that the overall formation constant for Fe(CN)6)-is ~1032, and that Fe3+ (aq) + e-= Fe2+ (aq) [Fe(CN)s! (aq) +e Fe(CN) +0.77v +0.36 (aq) E This the wite 39 answer λ 10
Write down the reduction reaction between Fe(CN)63- and Fe(CN)64-; find the standard electrode potential of this half reaction. Calculate the ionic strength of 0.1 KCl and 0.1 M Na2SO4, respectively.
3. When Clz(9) is added to aqueous Ke[Fe(CN).), [Fe(CN).]*- is oxidized to [Fe(CN).jº-and Kz[Fe(CN).] precipitates, as shown below. 2K [Fe(CN)6] (aq) +C12(g) = 2K [Fe(CN).](s) + 2Cl(aq) a. Write the net ionic equation for the reaction and calculate its equilibrium constant. (10 pts) At 25°C: [Fe(CN)6]3-(aq) + [Fe(CN)6]+-(aq) E° = 0.356 V Cl2(g) + 2e + 2Cl(aq) E° = 1.360 V Ksp of K3[Fe(CN)6] = 106.5
A solution is prepared by adding 0.075 mole of K3 [Fe(CN)6] to 0.72 L of 2.2 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. [Fe(CN)6*] = [Fe3+] = Submit Answer Try Another Version 2 item attempts remaining
A solution is prepared by adding 0.063 mole of K3 [Fe(CN)6] to 0.59 L of 2.1 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1.0 x 1042. M [Fe(CN)6*] = C [Fe3+]= C M Submit Anchor Another Voreion 2 itom attomnte romaining
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: b. When 80.0 mL of 1.00 M KCN is added to...
just 12 KEF CAND 372 C) de law is: a) Rate = k[Fe(CN)63-1*[I][Fe(CN)64 1°[12] Rate = k[Fe(CN).-1*[1][Fe(CN). [12] Rate = k[Fe(CN).-)]*[1] d) Rate = k[Fe(CN).-1[15] Rate = k[Fe(CN)63-1[1] [Fe(CN).-) Tabulated below are initial rate data for the reaction: 2Fe(CN).- + 21 2Fe(CN).-- + 12 12. [I do 0.01 0.02 0.02 0.02 0.02 [Fe(CN)64-) 0.01 0.01 0.01 3 [12] 0.01 0.01 0.01 0.01 0.02 Rate (M/s) 1x 10-5 2x 10-5 8 x 10-5 - 8 x 10-5 8 x 10-5 0.02...