Instrumental Analysis Electrogravimetry , Coulometry Tutorial 7

Objectives By the end of this session, the student should be able to: To calculate potential of a cell when current passes. Differentiate between electrogravimetry, coulometry. Correlate concentration to the electrochemical phenomena measured.

Ecell = Ecathode  Eanode  IR  OVERPOTENTIALS Example 1:Electrolytic cell Calculate the initial voltage that should be applied to electrolyze 0.010 M [Zn(OH)4]2 in 0.10 M NaOH, using Ni electrodes. Assume that the current is 0.20 A, the cell resistance is 0.35 Ω, and O2 is evolved at 0.20 bar. The over-potential for O2 evolution at a Ni surface at a current of 0.20 A is 0.519 V. The reactions are: Cathode: [Zn(OH)4]2- + 2e Zn(s) + 4OH E° = –1.199 V Anode: H2O 1/2O2 + 2H+ + 2e E° = 1.229 V Ecell = Ecathode  Eanode  IR  OVERPOTENTIALS Solution Written as reduction: 1/2O2 + 2H+ + 2e H2O [NaOH]= 0.1, pOH = 1 pH = 13, [H+] = 10-13

Example 2:Electrogravimetry Ions that react with Ag+ can be determined electro-gravimetrically by deposition on a silver working anode: Ag(s) + X  AgX(s) + e- What will be the final mass of a silver anode used to electrolyze 75.00 mL of 0.0238 M KSCN if the initial mass of the anode is 12.463 g. Solution 75.00 mL of 0.0238 M KSCN = 1.785 mmol of SCN- which gives 1.785 mmol of AgSCN, CONTAINING 0.1037 g of SCN-. Mass of SCN- = 1.785 mmol SCN- x 10-3 x 58.09 (g/mol SCN-) = 0.1037 g SCN- Final mass of silver anode = 12.4638 + 0.1037 = 12.5675 g

Theoretical Basis for Electrogravimetry and Coulometry Faraday’s law: The amount of chemical reaction at an electrode (mass of copper deposited on the cathode surface) is proportional to the quantity of electricity passed in the circuit. Faradaic current: The current passes in the circuit as a result of actual electrolysis (oxidation and reduction at electrode surface). Coulombs Amperes seconds If a reaction requires n electrons per mole of reactant, the quantity reacting of chemical species in time t is Faraday’s Law 5 5

Example 3: A 1.00-L electrolysis cell initially containing 0.0250 M Mn2+ and another metal ion, M3+, is fitted with Mn and Pt electrodes. The reactions are: Mn(s) → Mn2+ + 2e- M3+ + 3e- → M(s) a) Is the Mn electrode the anode or the cathode? b) A constant current of 2.60 A was passed through the cell for 18.0 min, causing 0.504 g of the metal M to plate out on the Pt electrode. What is the atomic mass of M? c) What will the concentration of Mn2+ in the cell be at the end of the experiment?

Solution Sine Mn is oxidized, it is the anode. b) Since 1 mol of M gives 3 e- Atomic mass of M = 0.504 g / 0.0097 mol = 52.0 g/mol One step solution: Wt. of M3+ Electrolyzed = It/nF X atomic.wt = 2.6X18X60/3X96500Xatomic.wt 0.504 = 0.0097 X atomic.wt atomic.wt = 52.0 g/mol c) In the electrolysis : no. of moles of Mn2+ produced = It/nF =2.6 X 18 X 60/2 X 96500 =0.01455mol Total moles of [Mn2+] = 0.0250 + 0.01455 = conc. Of [Mn2+] = 0.0396 moles in 1 Liter = 0.0396 M

Example 4: Coulometric titration H2S(aq) can be analyzed by titration with coulometrically generated I2. To 50.0 mL of sample were added 4 g KI. Electrolysis required 812 s at 52.6 mA. Calculate the concentration of H2S (g/mL) in the sample. Solution: Faraday's Law The number of grams reduced at the cathode or oxidized at the anode is given by: Where I = current in amps t = time in seconds FW = formula weight n = number of electrons transferred per 1 mole of species Concentration of H2S= Wt of sample /volume = 7525/50 = 150.5 g/mL

Example 5: Coulometric titration Chlorine has been used for decades to disinfect drinking water. An undesirable side effect of this treatment is the reaction of chlorine with organic impurities to create organochlorine compounds, some of which could be toxic. Monitoring total organic halide is now required for many water providers. A standard procedure is to pass water through activated charcoal that adsorbs organic compounds. Then charcoal is combusted to liberate hydrogen halide: Organic halide (RX) CO2 + H2O + HX The HX is absorbed into aqueous solution and measured by automatic coulometric titration with a silver anode: Ag(s)  Ag+ + e- (Ag+ generated anodically) X-(aq) + Ag+  AgX(s) (formed AgX is deposited on anode) When 1.00 L of drinking water was analyzed, a current of 4.23 mA was required for 387 s. A blank prepared by oxidizing charcoal required 6 s at 4.23 mA. Express TOX of the drinking water as mol halogen/L. If all halogen is chlorine, express the TOX as g/L

Solution - The corrected coulometric titration time is 387 – 6 = 381 s q = It/F = [(4.23 mA)(381 s)] / (96500 C/mol) = 0.0167 mmol e- = 16.7 mol e-. - Because 1e- is equivalent to one x-, the concentration of organohalide is 16.7 M. - If all halogen is Cl, this corresponds to 592 g Cl/L (16.7 mol/L x 35.45 g/mol). One Step solution Wt of Cl = It/nf X M.Wt = 4.23x 10-3 x 381/1x96500 x 35.45= 0.000592 gms = 592 g Conc of Cl= 592 per 1 liter sample = 592 g Cl/L