Given the following two half-reactions, write the overall reaction inthe direction in which it is spontaneous and calculate thestandard cell potential.Pb2+ (aq) + 2 e → Pb(s)Fe3*(aq) + e -→ Fe2*(s)E° = -0.126 VE° = +0.771 VPb2*(aq) + Fe?*(s) → Pb(s) + Fe³+(aq)8°.°cell= +0.645 VO → Pb2+(aq) + 2 Fe2*(s) ɛ°cell = +1.416 VPb(s) + 2 Fe3+(aq)O Pb(s) + Fe3+(aq) → Pb2*(aq) + Fe2* (s)ɛ°cell = +0.645 VO Pb(s) + 2 Fe3+(aq) → Pb2*(aq) + 2 Fe2* (s) ɛ°cell = +0.897 VO Pb2*(aq) + 2 Fe2* (s) → Pb(s) + 2 Fe3+(aq)ɛ°cell = +0.897 V

Answered: Image /qna-images/answer/3a2dd3dc-f173-4223-bd35-ca2abd1892ed.jpg

Given the following two half-reactions, write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e Fe3+(aq) + e -→ Pb(s) Fe2+(s) E° = -0.126 V E° = +0.771V → O Pb?*(aq) + Fe2*(s) → Pb(s) + Fe3+(aq) E°cell = +0.645 V O Pb(s) + 2 Fe3+(aq) → Pb2*(aq) + 2 Fe²*(s) ɛ°cell = +1.416 V O Pb(s) + Fe3+(aq) → Pb2*(aq) + Fe2*(s) E°cell = +0.645 V O Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe2* (s) ɛ°cell = +0.897 V O Pb2*(aq) + 2 Fe2*(s) → e°cell = +0.897 V Pb(s) + 2 Fe3+(aq)

Given the following two half-reactions, write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e Fe3+(aq) + e -→ Pb(s) Fe2+(s) E° = -0.126 V E° = +0.771V → O Pb?(aq) + Fe2(s) → Pb(s) + Fe3+(aq) E°cell = +0.645 V O Pb(s) + 2 Fe3+(aq) → Pb2(aq) + 2 Fe²(s) ɛ°cell = +1.416 V O Pb(s) + Fe3+(aq) → Pb2(aq) + Fe2(s) E°cell = +0.645 V O Pb(s) + 2 Fe³(aq) → Pb²(aq) + 2 Fe2* (s) ɛ°cell = +0.897 V O Pb2(aq) + 2 Fe2(s) → e°cell = +0.897 V Pb(s) + 2 Fe3+(aq)

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...

See similar textbooks

Similar questions

In the following: Al³+ (aq) + 3e → Al(s) E° = -1.66 V Cr³+ (aq) + 3e → Cr(s) E° = -0.74 V What is the standard cell potential for the corresponding galvanic cell? 0.00 V O-0.92 V +2.40 V +0.92 V O-2.40 V
The free energy change for the following reaction at 25 °C, when [H+] = 1.20 M and [Pb2+] = 3.36x10-3 M, is -38.9 kJ: 2H*(1.20 M) + Pb(s)→ H2(g) + Pb2+(3.36x10-3 M) AG = -38.9 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction v forward reverse
Given: Br2 (1) + 2e 2 Br -1 (aq); 12 (s) + 2e-2 1-¹ (aq); E 0= 0.54 V What is the standard cell potential for the following reaction? 12(s) + 2 Br ¹ (aq) → 21−¹(s) + Br2 (aq) O-1.63 V O-1.63 V O 0.87 V 0.55 V EO = 1.09 V -0.55 V
Use the tabulated half-cell potentials to calculate ΔG° for the following balanced redox reaction. I2(l)+2e- → 2I+(aq) E°cell = +0.54 Fe3+(aq)+e- → Fe2+(aq) E°cell = -0.04 3 I2(s) + 2 Fe(s) → 2 Fe3+(aq) + 6 I⁻(aq) A. -3.3 x 102 kJ B. +2.3 x 102 kJ C. -9.7 x 101 kJ D. +4.9 x 101 kJ E. -1.1 x 102 kJ
Use the tabulated half-cell potentials to calculate the standard cell potential for the following balanced redox reaction. 312(s)+2 Fe(s) → 2 Fe3+(aq) +61 (aq) 12(s)+2e 21-(aq) E°=+0.54 V Fe3+(aq) +3 e → Fe(s) E°=-0.04 V A) +0.29 V B) +0.50 V C) -0.50 V D) -0.58 V E) +0.58 V
The standard cell potential E°(cell) is +0.34 V for the redox reaction 2 In(s) + 6 H⁺(aq) → 2 In³⁺(aq) + 3 H₂(g). Determine E° for the half-reaction In³⁺(aq) + 3 e⁻ → In(s).
Calculate the [Ni2+] in the following galvanic cell, given that the measured voltage of the cell is 1.120 V at 25oC. Ni(s) | Ni2+(aq) ( ? M) || Ag+(aq) (0.84 M) | Ag(s) Ni2+(aq) + 2 e- → Ni(s) Eo = -0.257 V Ag+(aq) + e- → Ag(s) Eo = 0.800 V Your answer should have 2 significant figures.
The free energy change for the following reaction at 25 °C, when [H*] = 3.34x103 M and [Hg+] = 1.15 M, is 180. kJ: 2H*(3.34x103 M) + Hg(1)→ H2(g) + Hg²*(1.15 M) AG = 180. kJ What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction?
When the Ag+ concentration is 1.44 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 3.269 V. What is the Mg2+ concentration? 2Ag+ (aq) + Mg(s) → 2Ag(s) + Mg²+ (aq) red Ag+ (aq) + e¯ → Ag(s) Eº = 0.799 V Mg²+ (aq) + 2e¯ → Mg(s) Fre = -2.370 V Mg²+] = M
Chemistry Question
Calculate E°cell for the cell for the reaction 2 Ga(s) + 3 Sn4+(aq) → 3 Sn2+(aq) +2 Ga3+(aq).The standard reduction potentials are as follows: Ga3+(aq) + 3 e− → Ga(s) E° = –0.55 V Sn4+(aq) + 2 e− → Sn2+(aq) E° = 0.15V
A chemist designs a galvanic cell from the following two half-reactions NO3(aq) + 4H+ (aq) + 3e → NO(g) + EO red = +0.96 V 2H₂0 (1) Cu2+ (aq) + e → Cut (aq) Ere red= +0.153 V The standard cell potential for this galvanic cell is [Select] The standard free energy for this galvanic cell is AG [Select] The equilibrium constant (K) for this galvanic cell is [Select] V..