Fundamentals Of Analytical Chemistry
9th Edition
ISBN: 9781285640686
Author: Skoog
Publisher: Cengage
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- 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-.arrow_forwardWhat cathode potential (versus SCE) would be required to lower the total Hg(II) concentration of the following solutions to 1.00 10-6 M (assume reaction product in each case is elemental Hg): (a) an aqueous solution of Hg2+? (b) a solution with an equilibrium SCN- concentration of 0.100 M? Hg2+ + 2SCN- Hg(SCN)2(aq) = Kf = 1.8 107 c) a solution with an equilibrium Br- concentration of 0.100 M? HgBr42++ 2e- Hg(l) + 4Br- E0= 0.223 Varrow_forwardAn aqueous solution of an unknown salt of vanadium is electrolyzed by a current of 2.50 amps for 1.90 hours. The electroplating is carried out with an efficiency of 95.0%, resulting in a deposit of 2.850 g of vanadium. a How many faradays are required to deposit the vanadium? b What is the charge on the vanadium ions (based on your calculations)?arrow_forward
- 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.arrow_forwardThe saturated calomel electrode. abbreviated SCE. is often used as a reference electrode in making electrochemica1 measurements. The SCE is composed of mercury in contact with a saturated solution of calomel (Hg2Cl2). The electrolyte solution is saturated KCI. is +0.242 V relative to the standard hydrogen electrode. Calculate the potential for each of the following galvanic cells containing a saturated calomel electrode and the given half-cell components at standard conditions. In each case. indicate whether the SCE is the cathode or the anode. Standard reduction potentials are found in Table 17.1. a. Cu2++2eCu b. Fe3++e-Fe2+ c. AgCl+e-Ag+Cl- d. Al3++3eAl e. Ni2++2eNiarrow_forwardThe cell Ag|AgCl(sat’d)||H+(a = x)|glass electrode has a potential of —0.2094 V when the solution in the right-hand compartment is a buffer of pH 4.006. The following potentials are obtained when the buffer is replaced with unknowns: (a) —0.2806 V and (b) —0.2132 V. Calculate the pH and the hydrogen ion activity of each unknown. (c) Assuming anuncertainty of 0.001 Vin the junction potential, what is the range of hydrogen ion activities withinwhich the true value might be expected to lie?arrow_forward
- Calculate the voltages of the following cells at 25°C and under the following conditions: (a) Cu|Cu+(0.80M)Hg22+(0.10M)|Hg|Pt (b) Cr|Cr3+(0.615M)Ni2+(0.228M)|Niarrow_forwardCalculate the voltages of the following cells at 25°C and under the following conditions: (a) Zn|Zn2+(0.50M)Cd2+(0.020M)|Cd (b) Cu|Cu2+(0.0010M)H+(0.010M)|H2(1.00atm)|Ptarrow_forwardIn some old European churches, the stained-glass windowshave so darkened from corrosion and age that hardly anylight comes through. Microprobe analysis showed that tinycracks anddefects on the glass surface were enriched ininsoluble Mn(III) and Mn(IV) compounds. From AppendixE, suggest a reducing agent and conditions that might successfullyconvert thesecompounds to soluble Mn(II) withoutsimultaneously reducing Fe(III) (which gives the glassits colors) to Fe(II). Take MnO2 as representative of theinsoluble Mn(III) and Mn(IV) compounds.arrow_forward
- What is the cell potential of the following cell at 25C? Ni(s)Ni2+(1.0M)Sn2(1.5104M)Sn(s)arrow_forwardConsider only the species (at standard conditions) Na+, Cl, Ag+, Ag, Zn2+, Zn, Pb in answering the following questions. Give reasons for your answers. (Use data from Table 17-1.) a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by SO42 (aq) in acid? d. Which species can be reduced by Al(s)?arrow_forwardThe table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals A. B. C. D. and E. and their respective 1.00 M 2+ ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table 17-1 in the text. Assign a reduction potential of 0.00 V to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct. A(s)in A2+(aq) B(s)in B2+(aq) C(s)in V2+(aq) D(s)in D2+(aq) E(s)in E2+(aq) 0.28V 0.81V 0.13V 1.00V D(s)in D2+(aq) 0.72V 0.19V 1.13V C(s)in V2+(aq) 0.41V 0.94V B(s)in B2+(aq) 0.53Varrow_forward
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