Chloride ions are added to the AglAg* half-cell of the cellAg|Ag*l|Cu²*|Cu to precipitate AgCI (s). The cell voltage is thenmeasured to be +0.072 V at 298 K. If [Cu2+] = 1.0 M in the Cu2+|Cu halfcell, what is [Ag*] in the Ag|Ag* half-cell? (First step: You will need touse the data in Table 17-1 to calculate E°cell.) Give your answer totwo significant figures.Ecell = E°cell - (0.0257 V/n) In(Q)E'cell = E°ox + E°red cidic Solution, [H+] = 1MCIO-(aq) + H2O + 2e-Lit(aq) + e-K+(aq) + e-Ba2+(aq) + 2e-Ca2+(aq) + 2e-Na+(aq) + e-Mg2+(aq) + 2e-Al3+(aq) + 3e-Mn2+(aq) + 2eZn2+(aq) + 2e-Cr3+(aq) + 3e-Fe2+(aq) + 2e-Cr3+(aq) + e-Cd2+(aq) + 2e-PBSO4(s) + 2e-TI+(aq) + e-Co2+(aq) + 2e-Ni2+(aq) + 2e-Agl(s) + e-Sn2+(aq) + 2e-Pb2+(aq) + 2e-2H+(aq) + 2e-AgBr(s) + e-S(s) + 2H+(aq) + 2e-Sn4+(aq) + 2e-SO.2-(aq) + 4H*(aq) + 2e-Efed (V)Li(s)→K(s)Ba(s)→Ca(s)Na(s)→Mg(s)Al(s)→Mn(s)→Zn(s)Cr(s)Fe(s)Cr2+(aq)Cd(s)-3.040Lithium is the strongest reducin-2.936-2.906-2.869-2.714-2.357-1.68-1.182-0.762-0.744-0.409-0.408-0.402Pb(s) + SO,2-(aq)TI(s)Co(s)Ni(s)Ag(s) + 1-(aq)Sn(s)Pb(s)-0.356-0.336->-0.282-0.236-0.152-0.141-0.127H2(g)Ag(s) + Br(aq)H2S(aq)Sn2+(aq)→SO2(g) + 2H2OCu*(aq)Cu(s)Cu(s)21-(aq)Fe2+(aq)2Hg(1)>Ag(s)Hg2+(aq)NO(g) + 2H20→Au(s) + 4CI-(aq)2Br(aq)RO = strongest oxidiz0.000R = strongest reduc0.0730.1440.154Cu2+(aq) + e-Cu2+(aq) +.2e-Cut(aq) + e-.2(s) + 2e-Fe3+ (aq).+ e-Hg2 2+(aq) + 2e-Ag+(aq) + e-2H92+(aq) + 2e-NO3-(aq) + 4H+(aq) + 3e-AuCla-(aq) + 3e-Br2(1) + 2e-O2(g) + 4H+(aq) + 4e-MnO2(s) + 4H+(aq) + 2e-Cr20,2-(aq) + 14H+(aq) + 6e-Cl2(g) + 2e-CIO3-(aq) + 6H*(aq) + 5e¬Au3+(ag) + Зе-MnO4 (aq) + 8H+(aq) + 5e-PbO2(s) + SO,2-(aq) + 4H*(aq) + 2e-H2O2(aq) + 2H+(aq) + 2e-Co3+ (aq) + e-F2(g) + 2e-0.155->0.1610.3390.5180.5340.7690.7960.7990.9080.9641.0011.0772H201.2291.229→MN2+(aq) + 2H20→2Cr3+(aq) + 7H20→2CI-(aq)Cb(g) + 3H20Au(s)Mn2+(aq) + 4H2OPbSO4(s) + 2H2O→2H20Co2+(aq)2F-(aq)1.331.3601.458ban1.4981.5121.6871.7631.9532.889Fluorine is the strongest cBasic Solution, [OH-] = 1MEfed (V)-0.891→Fe(s) + 20H-(aq)→H2(g) + 20H-(aq)Fe(OH)2(s) + OH-(aq)-0.828Fe(OH)2(s) + 2e-2H20 + 2e-Fe(OH)3(s) + e-S(s) + 2e-NO3-(ag) + 2H2O + 3e-NO3 (ag) + H2O + 2e-ClO4 (ag) + H2O + 2eO>(g) + 2H20 + 4e-CIo3 (ag) + 3H20+6e-0.547-0.445Lithium and fluorine ardous materials to work-0.140NO(g) + 40H-(aq)NO2 (aq) + 20H-(aq)ClO, (aq) + 20H-(aq)→4OH-(aq)CI-(aq) + 60H-(aq)CI-(aq) + 20H-(aq)0.0040.3980.4010.6140.890STANDARD17.2Increasing strength of oxidizing agentsIncreasing strength of reducing agents

Nernst Equation for Single Electrode Potential Ecell = E0 – [RT/nF] ln Q Where, Ecell = cell…

Chloride ions are added to the Ag|Ag* half-cell of the cell Ag|Ag*||Cu2|Cu to precipitate AgCl (s). The cell voltage is then measured to be +0.072 V at 298 K. If [Cu²*] = 1.0 M in the Cu2+|Cu half cell, what is [Ag*] in the Ag|Ag* half-cell? (First step: You will need to use the data in Table 17-1 to calculate E'cell.) Give your answer to two significant figures. Ecell = E°cell - (0.0257 V/n) In(Q) E'cell = E'ox + E°red

Chloride ions are added to the Ag|Ag* half-cell of the cell Ag|Ag||Cu2|Cu to precipitate AgCl (s). The cell voltage is then measured to be +0.072 V at 298 K. If [Cu²] = 1.0 M in the Cu2+|Cu half cell, what is [Ag] in the Ag|Ag half-cell? (First step: You will need to use the data in Table 17-1 to calculate E'cell.) Give your answer to two significant figures. Ecell = E°cell - (0.0257 V/n) In(Q) E'cell = E'ox + E°red

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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