19eqeq2re2req2reqIts 2reqots 2req#3Use the References to access important values if needed for this question.Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:Cu2+ (aq) + Fe(s)→ Cu(s) + Fe²+ (aq)ECu2+ (aq) + 2eFe2+ (aq) +2eAG =kjK for this reaction would beSubmit Answer$4R→ Cu(s) Ee red→ Fe(s) Fred%5= 0.337 VRetry Entire Group 9 more group attempts remainingT= -0.440 Ve than one.Cengage Learning Cengage Technical Support6MacBook ProY887&U*81(9)0(Pre-Email Ins0 P

Answer:- This question is answered by using the simple concept of calculation standard Gibbs free…

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Cu²+ (aq) + Fe(s)→ Cu(s) + Fe²+ (aq) Cu2+ (aq) + 2e Fe2+ (aq) + 2e Submit Answer → Cu(s) Ee Cu(s) = 0.337 V red AG: kj K for this reaction would be → Fe(s) E Fe(s) E= -0.440 V red e than one. Retry Entire Group 9 more group attempts remaining

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Cu²+ (aq) + Fe(s)→ Cu(s) + Fe²+ (aq) Cu2+ (aq) + 2e Fe2+ (aq) + 2e Submit Answer → Cu(s) Ee Cu(s) = 0.337 V red AG: kj K for this reaction would be → Fe(s) E Fe(s) E= -0.440 V red e than one. Retry Entire Group 9 more group attempts remaining

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 standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Pb²+ (aq) + Cu(s) - → Pb(s) + Cu²+ (aq) AG° Pb²+ (aq) + 2e¯¯ → Pb(s) Ee Cu²+ (aq) + 2e¯ → Cu(s) Fre red = kJ K for this reaction would be = -0.126 V = 0.337 V + than one.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2Cu²+ u²+ (aq) + 21- (aq) → 2Cut (aq) + 12 (8) 2+ Cu²+ (aq) + e → Cut (aq) Ee red I2 (s) +2e → 21¯ (aq) AGO = K for this reaction would be kJ E red - 0.153 V 0.535 V 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
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Use standard reduction potentials to calculate the equilibrium constant for the reaction: Cu²+ (aq) + Pb(s) → Cu(s) + Pb²+ (aq) 2+ Cu²+ (aq) + 2e Pb²+ (aq) + 2e → Cu(s) Fre 0.337 V → Pb(s) Fe -0.126 V red = Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant at 298 K: AGO for this reaction would be | than zero.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Cu²+ (aq) + 2Cu+ (aq) → Cu(s) + 2Cu²+ (aq) 2+ Cu²+ (aq) + 2e → Cu(s) Ee = 0.337 V red 0.153 V = 2+ Cu²+ (aq) +e → Cut (aq) E red AG° kJ K for this reaction would be than one.
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