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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: Cd²+ (aq) + 2Fe²+ (aq) Cd(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: Cd²+ (aq) + 2Fe²+ (aq) Cd(s) + 2Fe³+ (aq) Answer: kj K for this reaction would be than one.

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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
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Standard Reduction (Electrode) Potentials at 25 °C Hg₂²*(aq) + 2 e F₂(g) + 2 e Ce¹(aq) + e Ce³+ (aq) MnO₂ (aq) +8 H (aq) +5 e Cl₂(g) + 2 e 2 Cl' (aq) Cr₂0₂² (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) + e Ag(s) Fe³+ (aq) +e 1₂ (5) + 2 e 2 F (aq) Half-Cell Reaction →→→ 2 Hg(!) →→→Fe²+ (aq) →21 (aq) Fe(CN),³(aq) + e Cu²+ (aq) + 2 e Cu²+ (aq) +e S(s) + 2 H* (aq) + 2 H₂S(aq) 2 H*(aq) + 2 e Pb²+(aq) +2 e Sn²+ (aq) + 2 e Ni²+ (aq) +2 e Co²+(aq) +2 e Cd²+ (aq) + 2 e Cr²³(aq) + e Fe²+ (aq) + 2 e Cr³+(aq) +3 e Zn²2(aq) + 2 e 2 H₂O(1) + 2 e > Fe (CN), ¹ (aq) Cu(s) Cu*(aq) H₂(g) > Pb(s) Sn(s) →→Ni(s) Co(s) Cd (s) Cr²2 (aq) Mn²+(aq) +2 e Al³+ (aq) + 3 e Mg2+ (aq) + 2 e Na (aq) +eNa(s) K(aq) + e →→→K(s) Li (aq) +eLi(s) Mn²+ (aq) + 4H₂O(1) →→Fe(s) →→Cr(s) Zn(s) > H₂(g) + 2 OH(aq) →→ Mn(s) →→ Al(s) Mg(s) Eº (volts) 2.87 1.61 1.51 1.36 1.33 1.229 1.08 0.96 0.920 0.855 0.799 0.789 0.771 0.535 0.48 0.337 0.153 0.14 0.0000 -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
Transcribed Image Text:Standard Reduction (Electrode) Potentials at 25 °C Hg₂²*(aq) + 2 e F₂(g) + 2 e Ce¹(aq) + e Ce³+ (aq) MnO₂ (aq) +8 H (aq) +5 e Cl₂(g) + 2 e 2 Cl' (aq) Cr₂0₂² (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) + e Ag(s) Fe³+ (aq) +e 1₂ (5) + 2 e 2 F (aq) Half-Cell Reaction →→→ 2 Hg(!) →→→Fe²+ (aq) →21 (aq) Fe(CN),³(aq) + e Cu²+ (aq) + 2 e Cu²+ (aq) +e S(s) + 2 H* (aq) + 2 H₂S(aq) 2 H*(aq) + 2 e Pb²+(aq) +2 e Sn²+ (aq) + 2 e Ni²+ (aq) +2 e Co²+(aq) +2 e Cd²+ (aq) + 2 e Cr²³(aq) + e Fe²+ (aq) + 2 e Cr³+(aq) +3 e Zn²2(aq) + 2 e 2 H₂O(1) + 2 e > Fe (CN), ¹ (aq) Cu(s) Cu*(aq) H₂(g) > Pb(s) Sn(s) →→Ni(s) Co(s) Cd (s) Cr²2 (aq) Mn²+(aq) +2 e Al³+ (aq) + 3 e Mg2+ (aq) + 2 e Na (aq) +eNa(s) K(aq) + e →→→K(s) Li (aq) +eLi(s) Mn²+ (aq) + 4H₂O(1) →→Fe(s) →→Cr(s) Zn(s) > H₂(g) + 2 OH(aq) →→ Mn(s) →→ Al(s) Mg(s) Eº (volts) 2.87 1.61 1.51 1.36 1.33 1.229 1.08 0.96 0.920 0.855 0.799 0.789 0.771 0.535 0.48 0.337 0.153 0.14 0.0000 -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
Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction: Cd²+(aq) + 2Fe2+(aq) Cd(s) + 2Fe³+(aq) Answer: kj K for this reaction would be Answer: 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) kj than one. K for this reaction would be than one.
Transcribed Image Text:Use the standard reduction potentials located in the 'Tables' linked above to calculate the standard free energy change in kJ for the reaction: Cd²+(aq) + 2Fe2+(aq) Cd(s) + 2Fe³+(aq) Answer: kj K for this reaction would be Answer: 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) kj than one. K for this reaction would be than one.
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