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Give the cell diagram for the reaction below using | and || when applicable.
Cd + 2 Fe3+ ---->Cd2+ + 2 Fe2+
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- The following cell was found to have a potential of —0.492 V: Ag|AgCl(sat’d)||HA(0.200 M),NaA(0.300 M)|H2(1.00 atm),Pt Calculate the dissociation constant of HA, neglecting the junction potential.Halide ions can he deposited at a silver anode, the reaction being Ag(s) + X- AgX(s) +e- Suppose that a cell was formed by immersing a silver anode in an analyte solution that was 0.0250 M Cl-,Br-, and I -ions and connecting the half-cell to a saturated calomel cathode via a salt bridge. (a) Which halide would form first and at what potential? Is the cell galvanic or electrolytic? (b) Could I- and Br- be separated quantitatively? (Take 1.00 l0-5 M as the criterion for quantitative removal of an ion.) If a separation is feasible, what range of cell potential could he used? (c) Repeat part (b) for I- and Cl-. (d) Repeat part (b) for Br- and Cl-.Give the cell diagram for the reaction below.Cd + 2 Fe3+ ---->Cd2+ + 2 Fe2+
- A 30.0 mL solution of 0.0180 M Rh3+ in 1 M HClO4 was titrated with 0.0900 M Ce4+ to give Rh6+ and Ce3+. Calculate the potential (vs. Ag | AgCl) at the equivalence point. At what volume of titrant will the potential be equal to the standard potential of the Ce half-reaction (ignoring the Ag | AgCl electrode)?Given a cell: 2 Ag+(aq) + Cu(s) ----> 2Ag (s) + Cu2+ (aq), E(cell) = 0.346 V, E0(cell) = E0cathode - E0anode = 0.800 V - 0.340 V = 0.460 V, I. calculate the concentration of Cu2+ given that of Ag+ is 1.0 x 10-5MGiven a cell: 2 Ag+(aq) + Cu(s) ----> 2Ag (s) + Cu2+ (aq), E(cell) = 0.346 V, E0(cell) = E0cathode - E0anode = 0.800 V - 0.340 V = 0.460 V, I. calculate the concentration of Cu2+ given that of Ag+ is 1.0 x 10-5M Cu2+M= Keq =
- Calculate the cell potential when 50.00 mL of 0.100 M Ag+ solution is titrated with 50.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0.197 V , E0 Ag+/Ag = 0.799 V, Kçç (AgCl) = 1.8 x10-8 A. 0.704v B. 0.507V C.0.238v D. 0.435VIn a polarographic experiment of a 60 mL of 0.08 M Cu2+solution, a limitingcurrent was left on for 15 minutes. If the average current during the time of theexperiment is 6.0 μA, what fraction of the copper is removed from the solution? TheFaraday constant is 96485 C/mol of electron.A 1.000 g sample containing chlorides, iodides and inert materials was treated with dilute nitric acid followed by AgNO3. A precipitate of AgCl (143.32) and AgI (234.77) was produced and weighs 0.9238 g. On heating in a current of Cl2, the AgI is converted to AgCl, and the resulting product weighs 0.7238 g. Find the percentage of a) NaI (149.89) and b) NaCl (58.44) in the sample
- A 50.00 mL of 0.100 M Fe(II) is to be analyzed by redox titration with suitable reagents For the redox titration of 50.00 mL of 0.100 M Fe(II) with 0.100 M Ce(IV) in1 M H2SO4, (i) What is the reaction involved? (ii) Explain how the potential at the equivalence point can be determined.Calculate the potential at the equivalence point as measured by a platinumindicator electrode against a standard hydrogen electrode.Write the half-reaction and calculate the electrode potential of1. a zinc electrode in 0.0400 M Zn(NO3)22. a zinc electrode in 0.0100 M NaOH and saturated with Zn(OH)2(s)3. a zinc electrode in 0.0100 M Zn(NH3)4 2+ and 0.220 M NH3 4. a platinum electrode in 0.0566 M K4Fe(CN)6 and 0.00813 M K3Fe(CN)65. a platinum electrode in a solution buffered at pH 6.00 and saturated withH2(g) at 1.00 atmA titration of 50.0 mL of 0.10 M Sn2+ with 0.2 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ using Pt and calomel electrodes. Assuming the standard potential for Sn2+/Sn4+ = 0.139V ; calomel, 0.241 V; Fe2+/Fe3+ = 0.732 V. Calculate the voltage (E) at 25 ml, 50 ml, and 70 ml.