Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:2H+ (aq) + Cd(s) → H₂ (s) + Cd²+ (aq)2H+ (aq) + 2eCd2+ (aq) + 2eH2(s) Ered= =0.000 VCd(s) Fre=-0.403 VredAG=kjK for this reaction would bethan one.

First, let's identify the reduction and oxidation half-reactions and their potentials:- Reduction…

Answered: Use standard reduction potentials to…

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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Given the half-reactions and their respective standard reduction potentials 1. Cu³+ + 2 e- + Cu+ E; = +1.28 V Cu²+ + e E; = +0.15 V E; = = +0.52 V 2. + Cu+ 3. Cu²+ + 2 e- → Cu(s) = +0.34 V 4. Cu* + e- Cu(s) E calculate the standard reduction potential for the reduction half-reaction of Cu(III) to Cu(II). Cu³+ + e Cu2+ E° = V
Given the reactions below: Reduction Reaction Reduction Potential Ni2+(aq) + 2 e- → Ni(s) -0.26 V Mn2+(aq) + 2e- → Mn(s) -1.19 V Au3+(aq) + 3e- → Au(s) 1.50 V Which combination will have the most thermodynamically favorable reaction with the reductionreaction: Cu2+ (aq) + 2 e- → Cu(s) (E° = 0.34 V)? (Cu2+ must be reduced) Support your answer with a calculation(s).
Use standard reduction potentials to calculate the equilibrium constant for the reaction: I₂ (s) + 2Fe²+ (aq). → 21¯ (s) + 2Fe³+ (aq) 0.535 V, E³+ /Fe²+ From the table of standard reduction potentials: E 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. 0.771 V
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 31₂ (s) + 2Cr(s) → 61¯(s) + 2Cr³+ (aq) I₂ (s) + 2e¯ → 21¯ (s) Cr³+ r³+ (aq) + 3e¯ → Cr(s) E red kJ AGO = Eº = = 0.535 V red = -0.740 V
What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Ag(s) and Mn2+(aq) to give Ag+(aq) and Mn(s) Use the reduction potential values for Ag+(aq) of +0.80 V and for Mn2+(aq) of -1.18 V
Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kj for the reaction: Zn²+ (aq) + 2Cr²+ (aq) → Zn(s) + 2Cr³+ (aq) Answer: KJ K for this reaction would be ✓ greater lace han one.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2+ Cd²+ (aq) + Zn(s) → Cd(s) + Zn²+ (aq) AG⁰ = Cd2+ (aq) +2e → Cd(s) E red Zn²+ (aq) +2e → Zn(s) E red K for this reaction would be greater less = -0.403 V -0.763 V kJ than one.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction:Pb2+(aq) + Ni(s) Pb(s) + Ni2+(aq)Answer: kJK for this reaction would be than one.
Just the last one. And use table if needed
Half-reaction E0 (V) Cr3+(aq) + 3e- → Cr(s) -0.74 Fe2+(aq) + 2e- → Fe(s) -0.440 Fe3+(aq) + e- → Fe2+(aq) +0.771 Sn4+(aq) + 2e- → Sn2+(aq) +0.154 Determine the standard cell potential (E0cell) for the voltaic cell for this reaction below based on the table above. 3Sn4+(aq) + 2Cr(s) → 2Cr3+(aq) + 3Sn2+(aq)
The following half-reaction °E = -2,07 V. AlF63-(aq) + 3e- à Al(s) + 6F-(aq) calculate the equilibrium constant at 25°C for this reaction: Al3+(aq) + 6F-(aq) ↔ AlF63-(aq) K = ?
Given the standard reduction potentials: Zn²+ (aq) + 2e-> Zn(s) Ag (aq) + e Ag(s) Which species is the best oxidizing agent? Zn(s) Ag(s) + Ag (aq) 2+ o Zn (aq) E =-0.76 V E = 0.80 V

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Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+ (aq) + Cd(s) → H₂ (s) + Cd²+ (aq) 2H+ (aq) + 2e Cd2+ (aq) + 2e H2(s) Ered = =0.000 V Cd(s) Fre = -0.403 V red AG= kj K for this reaction would be than one.