Question 2 (1 point) Given the following half reactions with their corresponding standard reduction potentials, which...
Question 2 (1 point) Given the following half reactions with their corresponding standard reduction potentials, which of the following is/are true for the overall reaction happening under standard conditions? NO3 + 2H+ + 2e → NO2 + H20 E'O=0.421 V + O2 + 2H+ + 2e H20 E'=0.816 V (CHECK THE ONE(S) THAT IS/ARE CORRECT) 1) AE'' = 0.395 V for the spontaneous reaction. 2) % O2 + NO2 + NO3 is a spontaneous reaction. 3) O2 is the reducing...
Consider the following standard reduction potentials in acid solution:E^o(V)Al3+ + 3e– ? Al(s) –1.66AgBr(s) + e– ? Ag(s) + Br– +0.07Sn4+ + 2e– ? Sn2+ +0.14Fe3+ + e– ? Fe2+ +0.77The strongest reducing agent among those shown above is
pls explain in great detail why Consider the following standard reduction potentials Reduction Half-Reaction A1+ (aq) + 3e A1(s) Fe2+ (aq) + 2e-Fe(s) Sn2+ (aq) + 2e - Sn(s) E (volts) -1.66 0.44 0.14 The AIAP half-reaction can be paired with the other two to produce voltaic cells because Als* is a more powerful reducing agent OOO Al is a more powerful oxidizing agent o Fe and Sn are readily oxidized Al is a more powerful reducing agent AB+ is...
12. Using two half reactions that have NEGATIVE standard reduction potentials results results in a battery that... Reduction Half-Reaction F2(g) + 2e →2F(aq) S2082 (aq) + 2e- → 25042 (aq) O2(g) + 4H+ (aq) + 4e → 2H2O(1) Br2(1) + 2e + 2Br (aq) Agt(aq) + e → Ag(s) Fe3+ (aq) + e- → Fe2+ (aq) 126) + 2e → 21 (aq) Cu2+ (aq) + 2e → Cu(s) Sn4+ (aq) + 2e → Sn2+ (aq) S(s) + 2H+ (aq) +...
The following two half-reactions are found in a table of standard reduction potentials: Half-Reaction E cytochrome c1(Fe3+) +e- → cytochrome c1 (Fe2+) 0.22 V lipoic acid + 2 H+ + 2 e- → dihydrolipoic acid -0.29 V + What is value of Eº for the reaction that will occur involving these two half-reactions that will be spontaneous under standard conditions? 0 0.73 V 0 0.07 V O 0.51 V -0.07 V
Q7) Using Table 9.1 (page 294) in your textbook and/or in the slides of chapter 9 (online material), determine the standard free energy (AG) for the following reaction in kJ/mol. [Faraday constant = 96.5 kJ/V] [10 points) FADH2 + 1/202 - FAD + 2H+ + H:0 Show detailed calculation. Final answer without clear work will not be considered. TABLE 9.1 Standard Reduction Potentials Redox Half-Reaction 2H+ + 2e" - H a-Ketoglutarate + CO, + 2H+ 2e isocitrate NADP+ + H+...
consider the following standard reduction potentials. Reduction Half-Reaction Eo (volts) Al3+(aq) + 3e− → Al(s) − 1.66 Fe2+(aq) + 2e− → Fe(s) − 0.44 Sn2+(aq) + 2e− → Sn(s) − 0.14 The Al/Al3+ half-reaction can be paired with the other two to produce voltaic cells because ________ A) Al is a more powerful oxidizing agent B) Fe and Sn are readily oxidized Al is a more powerful reducing agent C) Al3+ is a more powerful oxidizing agent D) Al3+...
For the following reactions, determine E°, ΔG°, and K, given the balanced half reactions and the standard reduction potentials. Also, determine if the reaction is spontaneous as written. A) 2 Co3++ H3AsO3+ H2O 2 Co2++ H3AsO4+ 2H+ Co3++ e- Co2+ E°= 1.920 V H3AsO4+ 2H++ 2 e-H3AsO3+ H2O E°= 0.575 V Answer: (E° = 1.345 V; K = 2.95 x 1045; ΔG° = -2.60 x 105J; spontaneous) B) 4 Fe3++ 2 H2O 4Fe2++ O2+ 4 H+ Fe3++ e- Fe2+ E°= 0.771 V ½ O2+ 2...
Using the standard reduction potentials given below, decide which of the following reactions will occur spontaneously as written. Fe3+(aq) + e + Fe2+(aq) E° = 0.77 V Sn4+ (aq) + 2e → Sn2+(aq) E° = 0.13V Zn2+(aq) + 2e → Zn(s) E° = -0.77 V Lit(aq) + e + Li(s) E° = -3.05 V 2Li+ (aq) + Sn2+(aq) → 2Li(s) + Sn4+(aq) Sn2+(aq) + Zn2+(aq) → Sn4+(aq) + Zn(s) Lit(aq) + Fe2+(aq) - Li(s) + Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) →...
Given the following list of half-reaction reduction potentials, identify the reaction that will occur spontaneously as written: Half-reaction E° (V) -0.74 Cr3+ (aq) + 3 e ---> Cr (s) Sn4+ (aq) + 2 e ---> Sn2+ (aq) +0.154 -0.440 Fe2+ (aq) + 2 e ---> Fe(s) Fe3+ (aq) + e ---> Fe2+ (aq) +0.771 2 Cr (s) + 3 Fe2+ (aq) ---> 3 Fe (s) + 2 Cr3+ (aq) 2 Cr3+ (aq) + 3 Sn2+ (aq) ---> 3 Sn4+ (aq)...