Based on the half-reactions and their respective standard reduction potentials below, which element or ion will reduce an aqueous solution containing Fe3 (aq)? O2(s) + 4H* + 4e 2H20(1) 1.23 V Fe3 (aq) + e→ Fe2*(aq) 0.77 V Cu2*(aq) + 2e -→ Cu(s) 0.34 V 2H*(aq) + 2e → H2lg) 0.00 V Ni2*(aq) + + 2e Ni(s) -0.26 V Fe2*(aq) + 2e→ Fe(s) -0.45 V O aqueous copper(Il) acetate O nickel wire (solid nickel) O oxygen gas O none of these

Based on the half-reactions and their respective standard reduction potentials below, which element or ion will reduce an aqueous solution containing Fe3 (aq)? O2(s) + 4H* + 4e 2H20(1) 1.23 V Fe3 (aq) + e→ Fe2(aq) 0.77 V Cu2(aq) + 2e -→ Cu(s) 0.34 V 2H(aq) + 2e → H2lg) 0.00 V Ni2(aq) + + 2e Ni(s) -0.26 V Fe2*(aq) + 2e→ Fe(s) -0.45 V O aqueous copper(Il) acetate O nickel wire (solid nickel) O oxygen gas O none of these

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

Identify the strongest reducing agent based on the following half- reactions. 2+ +1.22 V → Mn (aq) + 2H₂O(1) +0.61 V MnO₂(s) + 4H* (aq) + 2e- Hg₂SO4(s) + 2e → 2Hg(1) + SO4 (aq) SnO₂(s) + 2H₂O0+ 4e → Sn(s) + 4OH(aq) Cr(OH)3(s) + 3e → Cr(s) + 3OH(aq) -0.95 V -1.48 V Cr Mn02 Hg2SO4 Sn Hg
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Round your answer to 2 significant digits. 2Cr(OH)3 (s) +100H¯ (aq) +3Zn²+ (aq) → 2CrO²¯ (aq) +8H₂O (1) + 3 Zn (s) 0 x10 X
Give only typing answer with explanation and conclusion
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy for the following redox reaction. Round your answer to 3 significant digits. The final units should be in kj
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction |HNO₂(aq)+H*(aq)+e¯ - →>> NO(g) + H2O(l) Fe3+(aq)+e → Fe (aq) Fe2+(aq) Answer the following questions about this cell. standard reduction potential +0 Ε = +1.00 V 0 red Ered = +0.771 V
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: (aq) + 2Cu+ (aq) → Pb(s) + 2Cu²+ (aq) AG° Pb²+ = Pb²+ (aq) + 2e¯ → Pb(s) Ee = -0.126 V red 2+ Cu²+ (aq) + e → Cu+ (aq) Fº red kJ K for this reaction would be = = 0.153 V + than one.
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG° for the following redox reaction. Round your answer to 3 significant digits. 6CI (aq)+2Cr0 (aq)+8H,0 (1) → 3Cl, (g)+2Cr(OH), (s) + 10OH (aq)
Calculate Eu for the following cell: Pt(s) | H₂(g) | H *(aq) || Pb2+ (aq) | PbSO4(s) | Pb(s) given the following standard reduction potentials. 2 H(aq) + 2e →→ H₂(g) PbSO4(s) + 2e →→ Pb(s) + SO42 (aq) O a. -0.712 V O b. +0.356 V O c. -0.356 V O d. +0.712 V O e. -0.178 V E° = 0.000 V E = -0.356 V
Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V
Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e- - → Pb(s) E° = -0.126 V Fe3*(aq) + e¯ → Fe2*(s) E° = +0.771 V Pb2*(aq) + 2 Fe²*(s) → Pb(s) + 2 Fe³*(aq); = +0.897 V cell Pb2*(aq) + Fe²*(s) → Pb(s) + Fe³*(aq); E, cell = +0.645 V Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); F° cell = +1.416 V Pb(s) + 2 Fe*(aq) → Pb²*(aq) + 2 Fe²*(s); E. 'cell = +0.897 V Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); = +0.645 V
Consider the following standard reduction potentials: Zn2*(aq) + 2 e → Zn(s) E° = -0.76 V Mg2*(aq) + 2 e → Mg(s) E° = -2.37 V Ag*(aq) + e → Ag(s) E° = +0.80 V Which is the strongest oxidizing agent? Mg2*(aq) Zn2*(aq) Ag*(aq) 5:08 PM 90°F Partly sunny P Type here to search 8/6/2021
Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy AG for the following redox reaction. Round your answer to 2 significant digits. H₂(g) + 2OH(aq) + Zn²+ (aq) → 2H₂O (1) + Zn (s) kJ X S ? 1