Consider the following half-reactions: Half-reaction E° (V) Ag*(aq) + e". » Ag(s) 0.799V Pb2*(aq) + 2e → Pb(s) -0.126V 2+ Zn2*(aq) + 2e → Zn(s) |-0.763V (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Ag(s) reduce Zn²*(aq) to Zn(s)? 2+

Consider the following half-reactions: Half-reaction E° (V) Ag(aq) + e". » Ag(s) 0.799V Pb2(aq) + 2e → Pb(s) -0.126V 2+ Zn2(aq) + 2e → Zn(s) |-0.763V (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Ag(s) reduce Zn²(aq) to Zn(s)? 2+

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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Consider the following half-reactions: Half-reaction E° (V) Ag*(aq) + e" Co2*(aq) + 2e → Ag(s) 0.799V → Co(s) |-0.280V Zn2*(aq) + 2e → Zn(s) -0.763V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Zn2*(aq) oxidize Ag(s) to Ag"(aq)? (6) Which species can be reduced by Co(s)? If none, leave box blank.
Consider the following unbalanced oxidation-reductionreactions in aqueous solution:Ag+(aq) + Li(s)----> Ag(s) + Li+(aq)Fe(s) + Na+(aq)----> Fe2+ (aq) + Na(s)K(s) + H2O(l)---->KOH(aq) + H2(g)(a) Balance each of the reactions. (b) By using data inAppendix C, calculate ΔH° for each of the reactions.(c) Based on the values you obtain for ΔH°, which of the reactionswould you expect to be thermodynamically favored?(d) Use the activity series to predict which of these reactionsshould occur. Are these results in accordwith your conclusion in part (c) of this problem?
3.2 calculate the standard cell potential of the following cell at 25C Sn(s)|Sn"(aq)||Cu?²*(aq)|Cu(s)
Consider the following cell. Ag(s)|AgCI(s)|NaCI(aq)|Hg2Cl2(s)|Hg(/)|Pt(s) (a) Write the half-cell reactions. (Include states-of-matter under the given conditions in your answer.) Cathode: chemPad Help Greek - Anode: chemPad O Help X.X"| Greek - (b) The standard emfs of the cell at several temperatures are as follows. T/K 291 298 303 311 E°/mV 43.0 45.4 47.1 50.1 Calculate the values of A,G°, A,Sº, and A,H° for the reaction at 303 K. A,G°: A,S° kJ•mol-1 J-mol -1.к-1 %3D A,H° kJ•mol-1
Consider the following half-reactions: Half-reaction E° (V) 2+ Hg"(aq) + 2e → Hg(1) 0.855V Co**(aq) + 2e → Co(s) -0.280V 2+ Fe"(aq) + 2e → Fe(s) |-0.440v (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Hg*(aq) oxidize Fe(s) to Fe2*(aq)?| (6) Which species can be oxidized by Co2*(aq)? If none, leave box blank.
Consider the following half-reactions: Half-reaction E° (V) Cl2(g) + 2e" → 2CI'(aq) 1.360v Ni?*(aq) + 2e- Ni(s) |-0.250V Mn2*(aq) + 2e → Mn(s) -1.180V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Mn2*(aq) oxidize Cl'(aq) to Cl2(g)? (6) Which species can be reduced by Ni(s)? If none, leave box blank.
R = 8.314 mol·K F = 96,485 mot AG = AG° + RT · In(Q) ΔΕΔΕ-).In(O) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (1) 1.229 2 z" (aq) Z2 (s) + 2 e 3+ (aq) + 3 е 0.426 A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e H2 (g) + 2 OH (aq) - 0.828 G (s) - 1.245 > M2+ (aq) + 2 e - 1.893 → M (s) A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. Calculate AG° in kJ/mol for this galvanic cell. Report your answer with 4 significant figures. You do not need to report units with your answer. If your value is negative, make sure to include a "-" symbol.
Consider the following half-reactions: Half-reaction E° (V) F2(g) + 2e → 2F*(aq) 2.870V Co2+*(aq) + 2e → Co(s) |-0.28OV 2+ Mg"(aq) + 2e → Mg(s) -2.37OV (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Mg(s) reduce F2(g) to F"(aq)? (6) Which species can be reduced by Co(s)? If none, leave box blank.
Consider the following half-reactions: Half-reaction ° (V) Br2(1) + 2e → 2Br°(aq) |1.080V Sn²+ (aq) + 2e . Sn(s) |-0.140V Zn2*(aq) + 2e → Zn(s)|-0.763V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Br2(0) oxidize Zn(s) to Zn2*(aq)? (6) Which species can be oxidized by Sn2*(aq)? If none, leave box blank.
Examine the following half-reactions and select the strongest oxidizing agent among the substances. v Jersey Departm E° = 0.755 V [PtCl4]²¯(aq) + 2e¯ 2 Pt(s) + 4CI'(aq); RuO4(s) + 8H*(aq) + 8e¯ 2 Ru(s) + 4H2O(1); FeO4²(aq) + 8H*(aq) + 3e¯ 2 Fe3*(aq) + 4H2O(/); H4XEO6(aq) + 2H*(aq) + 2e¯ 2 XeO3(aq) + 3H2O(/); E° = 2.42 V E° = 1.038 V E° = 2.07 V RuO4(s) H4XEO6(aq) [PtCl4]² (aq) O HFEO4¯ (aq) CI(aq)
5. A more complex example where other atoms are involved. Balance the following reaction in basic solution. r (aq) + MnO4 (aq) MnO, (s)+ BrO;" (aq) TABLE 19.1 Standard Electrode Potentials at 25 °C Reduction Half-Reaction E'(V) F + 20 H,0,(aq) + 2 H(aq)+2 e 2F(aq) 2.87 Stronger oxidizing agent Weaker 2 H,0 - PBSO (s) +2 H,0) -Mn0(s)+ 2 H,0) Mn (aq) 4 H,O) - Auts) 1.78 reducing agent PO,(s) + 4 H"(ag) + S0,2-(aq) + 2 e 1.69 Mno, (aq)+ 4 H'(aq) 3 e Mno, (aa) + 8 H (aq)+5 e Au (aq)+3e PbO,(s)+ 4 H(aa) + 2 e Cl,(g) + 2 e" 1.68 1.51 1.50 1.46 - P (ag) + 2 HO - 2a(aq) 1.36 1.33 Cr0, (aq) + 14 H'(ag) + 6 e 0+ 4 H (aq) + 4 e MnO(s)+4 H'(aq) +2 e 10, (aq) 6 H'(aq) + 5e Br+2e vo, (aq) + 2 H'(aq) + e NO, (aq) + 4 H°(aq) + 3e CIO+e 2 C"(aq) + 7 H,On - 2 H,O1) 1.23 Mn (aq) + 2 HyOV) 121 ulae) + 3 H,0() 2 Br (ag) - vo"(aq) +H,00 - NO + 2 H,O0 - CIO, (aa) - Agis) 1.20 1.09 1.00 0.96 0.95 Ag"(aa) + e 0.80 Fe (aq) + O+2 H'(aq)+2e - Fe*(aq) 0.77 - H,O,(ag) 0.70 Mn0, Taa) + e Mn0 (aa) 0.56 8) + 2 e…
here is the full question. can you please please not reject it and answer