Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction:Fe2+ (aq) + 2Fe2+ (aq) Fe(s) + 2Fe³+ (aq)Answer:kjK for this reaction would bethan one. Standard Reduction (Electrode) Potentials at 25 °CHalf-Cell Reaction2 F (aq)F₂(g) + 2 eCe¹(aq) + eMnO₂ (aq) + 8Cl₂(g) + 2 eCr₂0₂² (aq) + 14 H(aq) + 6 e —> 2 Cr³(aq) + 7 H₂0 (1)O₂(g) + 4 H* (aq) + 4e-2 H₂0 (1)Br₂(1) +2 e 2 Br (aq)NO₂ (aq) + 4 H*(aq) + 3 e° ——» NO(g) + 2 H₂0 (1)2 Hg2+ (aq) + 2 eHg₂² (aq)Hg2+ (aq) + 2 eHg(1)Ag (aq) + eAg(s)Hg₂²*(aq) + 2 eCe³+ (aq)H(aq) + 5 e2 Cl(aq)→→→>2 Hg(!)Fe³+ (aq) + e1₂ (s) + 2 e →21 (aq)→→→Fe²+ (aq)Fe(CN),³(aq) + e > Fe(CN),* (aq)Cu²+ (aq) + 2 eCu(s)Cu²+ (aq) +Cu(aq)S(s) + 2 H¹ (aq) + 2 H₂S(aq)2 H*(aq) + 2ePb²+ (aq) + 2 eSn²(aq) + 2 eNi²+ (aq) + 2 eCo²+ (aq) + 2 eCd²+ (aq) + 2 eCr²³(aq) + eFe²+ (aq) + 2 eCr³+ (aq) + 3 eZn²2(aq) + 2 e2 H₂O(1) + 2 eH₂(g)> Pb(s)Sn(s)→→Ni(s)Co(s)Cd (s)Cr²2 (aq)Mn²+ (aq) + 4H₂O(1)→→Fe(s)→→Cr(s)Zn(s)> H₂(g) + 2 OH(aq)Mn²+ (aq) + 2e →→→Mn(s)Al³+ (aq) + 3 e →→→> Al(s)Mg2+ (aq) + 2 eMg(s)Na (aq) +eNa(s)K(aq) + e →→→K(s)Li (aq) +eLi(s)Eº (volts)2.871.611.511.361.331.2291.080.960.9200.8550.7990.7890.7710.5350.480.3370.1530.140.0000-0.126-0.14-0.25-0.28-0.403-0.41-0.44-0.74-0.763-0.83-1.18-1.66-2.37-2.714-2.925-3.045

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Use the standard reduction potentials located in the Tables' linked above to calculate the standard free energy change in kJ for the reaction: Fe2+ (aq) + 2Fe2+ (aq) Fe(s) + 2Fe3+ (aq) Answer: kj K for this reaction would be than one.

Use the standard reduction potentials located in the Tables' linked above to calculate the standard free energy change in kJ for the reaction: Fe2+ (aq) + 2Fe2+ (aq) Fe(s) + 2Fe3+ (aq) Answer: kj K for this reaction would be than one.

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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Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction: Ni?+ (aq) + Cd(s) → Ni(s) + Cd²+ (aq) Answer: kJ K for this reaction would be than one.
Calculate the cell potential, E, for the given reactions at 25.00 "C using the ion concentrations provided. Then, determine if th cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E). Pt(s) + Fe²(aq) = P² (aq) + Fe(s) [Fe²*1 = 0.0064 M E= incorrect -4761 E= Cu(s) + 2 Ag* (aq) = Cu² (aq) + 2 Ag(s) |Cu²+1 = 0.027 M Ag1=0.027 M E [P1²*1 = 0.067 M Co² (aq) + Ti (aq) — Co¹ (aq) + Ti² '(aq) Co²1 = 0.055 M Co= 0.030 M ITP 1 = 0.0070 M T² = 0.0107 M mam V The cell is not spontaneous spontaneous. The cell is spontaneous not spontaneous. incorrect The cell is spontaneous not spontaneoux.
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Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr²+] = 0.768 M and [Fe2+] = 0.0170 M. Use the standard reduction potentials in this table. Cr(s) + Fe²+(aq) Cr² (aq) + Fe(s) E = x10 TOOLS V
Use the standard reduction potentials given below to predict if a reaction will occur when Mg metal is put into a 1 M aqueous Cd²+ solution. Cd(s) Eº = -0.403 V red Eº → Mg(s) Fred = -2.370 V Cd2+ (aq) + 2e → 2+ Mg²+ (aq) + 2e If a reaction will occur, write a balanced net ionic equation for the reaction. If no reaction will occur, leave all boxes blank. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) +
Use standard reduction potentials to calculate the equilibrium constant for the reaction:Co2+(aq) + Cu(s) Co(s) + Cu2+(aq)Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.Equilibrium constant: G° for this reaction would be than zero.
Consider the following standard reduction potentials: F2(g) + 2 e - 2 F (aq) E° = +2.87 V Mg2*(aq) + 2 e - Mg(s) E° = -2.37 V Under standard conditions, O Mg(s) is a stronger oxidizing agent than F (aq), and Mg2*(aq) is a stronger reducing agent than F2(g), O F (aq) is a stronger oxidizing agent than Mg(s), and F2(g) is a stronger reducing agent than Mg2*(aq). O F2(g) is a stronger oxidizing agent than Mg-"(aq), and Mg(s) is a stronger reducing agent than F (aq). O Mg2*(aq) is a stronger oxidizing agent than F2(g), and F(aq) is a stronger reducing agent than Mg(s).
Use the standard reduction potentials located in the "Tables' linked above to calculate the standard free energy change in kJ for the reaction: 3Hg"(aq) + 2Cr(s)3Hg(1) + 2Cr*(aq) Answer: |kJ K for this reaction would be than one.
Use the standard reduction potentials located in the Tables' linked above to calculate the standard free energy change in kj for the reaction: Fe²+ (aq) + Hg(1)→ Fe(s) + Hg²+ (aq) Answer: 237.16 K for this reaction would be greater than one.
What is the equilibrium constant for the following reaction at 25 °C? Round your answer to 1 decimal place. 2 Cu(s) + 2 Ag+ (aq) ⇒ Cu²+ (aq) + 2 Ag (s) = Note: Reference the Standard reduction potentials at 25 °C table for additional information. K = = x10
Calculate DeltaG (part 1 and 2)