Download Cr3+ + NO [acidic] and more Lecture notes Chemistry in PDF only on Docsity!
- Balance the following redox reactions using the half-reaction method. (10 pts)
a. Cr + NO 3 -^ → Cr3+^ + NO [acidic]
Cr → Cr3+^ + 3e- 3e-^ + 4H+^ + NO 3 -^ → NO + 2H 2 O 4H+^ + Cr + NO 3 -^ → Cr3+^ + NO + 2H 2 O
b. CrO 4 2-^ + SO 3 2-^ → Cr(OH) 3 + SO 4 2-^ [basic]
2 x (3e-^ + 5H+^ + CrO 4 2-^ → Cr(OH) 3 + H 2 O) 3 x (H 2 O + SO 3 2-^ → SO 4 2-^ + 2H+^ + 2e-) 10H+^ + 2CrO 4 2-^ + 3H 2 O + 3SO 3 2-^ → 2Cr(OH) 3 + 2H 2 O + 3SO 4 2-^ + 6H+ 4 10H+^ + 2CrO 4 2-^ + 3H 2 O + 3SO 3 2-^ → 2Cr(OH) 3 + 2H 2 O + 3SO 4 2-^ + 6H+ 4OH-^ 4OH- 2CrO 4 2-^ + 5H 2 O + 3SO 3 2-^ → 2Cr(OH) 3 + 3SO 4 2-^ + 4OH-
- Calculate Ecell and ΔG for the following voltaic cell under non-standard conditions: [Cu2+] = 0.10 M and [Zn2+] = 1.90 M. (10 pts)
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) Nernst Equation
𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙°^ −
F = 96500 J/Vmol
E°cell = Ecat – Ean = 0.342 V – (-0.762 V) = 1.104 V
𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑐𝑒𝑙𝑙°^ −
ΔG = -nFE = -(2 mol e-)(96500 J/V mol)(1.066 V) = -205767 J or -205.8 kJ
Standard Reduction Potentials E° 2e- + Cu2+^ → Cu 0.342 V 2e- + Zn2+^ → Zn - 0.762 V
- Complete the following nuclear reactions. Clearly state the nuclear decay mode to receive full credit. (10 pts) Negatron Emission Electron Capture
a) 2760 Co (^) 10 e + 2860 𝑁𝑖^ b) 9743 Tc + (^) −1^0 𝑒^ 4297 Mo + 00 𝛾
Alpha Particle Decay Positron Emission
c) 22688 Ra 24 He + 22286 𝑅𝑛^ d) 20583 Bi 10 e + 20582 𝑃𝑏
- Using the spectrochemical series, fill the electrons in the following crystal field diagram for the following diamagnetic transition metal complex. Is this the ligand a strong field ligand or a weak field ligand? Also predict if the compound is square planar or tetrahedral. (10 pts)
Spectrochemical Series I-<Br-<Cl-<F-<OH-<H 2 O//<NH 3 <en< NO 2 - <CN-<CO Weak Field ligands (High Spin) Strong Field Ligands (Low Spin) Tetrahedral Square Planar [Ni(CO) 4 ]2+ CO gives low spin = strong field _____ Ni2+^ dx^2 -y^2 d^8 ion ↑↓ ↑ ↑ ↑↓ dxy dyz dxz dxy
↑↓ ↑↓ ↑↓ dz^2 dz^2 dx^2 -y^2 Paramagnetic ↑↓ ↑↓ dxz dyz Compound is Square planar. Diamagnetic
- Using the short-hand notation (Noble Gas Configuration), write the ground-state electron configuration for the following transition metals: (10 pts)
a. Fe2+^ b. Re7+^ c. Cu1+^ d. Pd2+
[Ar]3d^6 [Xe]4f^14 [Ar]3d^10 [Kr]4d^8
- The complex ion [Ti(H 2 O) 6 ]3+^ is violet in solution. Calculate the crystal field splitting energy in kJ/mol for this ion. (10 pts) h = 6.626 x 10-34^ Js c = 3.00 x 10^8 m/s Absorbed wavelength (nm) Observed Color 6.02 x 10^23 ions/mol 400 violet greenish-yellow 450 blue yellow 1 nm = 10-9^ m 490 blue-green red 570 yellow-green violet 580 yellow dark-blue 600 orange blue 650 red green
∆ = 𝐸 =
ℎ𝑐 𝜆 =
(6.626𝑥10−34𝐽 𝑠)(3.00𝑥
(^8) 𝑚 𝑠 ) (570 𝑛𝑚)(^10
−9𝑚 𝑛𝑚 )
×
6.02𝑥10^23 𝑖𝑜𝑛/𝑚𝑜𝑙 1000 𝐽/𝑘𝐽 = 𝟐𝟏𝟎 𝒌𝑱/𝒎𝒐𝒍
- Write a partial decay series for Ra-226 undergoing the following sequential decays. (10pts)
Alpha → Alpha → negatron → alpha Alpha Decay (^22688) 𝑅𝑎 (^) → 24 𝐻𝑒 (^) + 22286 𝑅𝑛
Alpha Decay (^22286) 𝑅𝑛 (^) → 24 𝐻𝑒 (^) + 21884 𝑃𝑜
Negatron Emission (^21884) 𝑃𝑜 (^) → (^) −1 (^0) 𝑒 (^) + 21885 𝐴𝑡
Alpha Decay (^21885) 𝐴𝑡 (^) → 24 𝐻𝑒 (^) + 21483 𝐵𝑖
- Extra Credit. The following solutions of nickel absorb light in the following regions of the electromagnetic spectrum: NiCl 4 2-(702 nm) and NiBr 4 2-(756 nm). Which ion has the greater crystal field splitting energy? (5 pts)
𝐶𝑟𝑦𝑠𝑡𝑎𝑙 𝐹𝑖𝑒𝑙𝑑 𝑆𝑝𝑙𝑖𝑡𝑡𝑖𝑛𝑔 𝐸𝑛𝑒𝑟𝑔𝑦 = ∆= 𝐸 =
E702 nm > E756 nm
