Chemistry very hard take home problem please provide bolded and clear and correct anwsers thank you...
Use the chart to answer the questions. Please be correct and careful In one step of glycolysis, glyceraldehyde 3-phosphate is oxidized by NAD+ to yield 3-phosphoglycerate and NADH. 2 2 2 Table 10-2 Standard Reduction Potentials for Redox Pairs of Biological Relevance* Redox Pair Number of (oxidized form → reduced form) Electrons E.(V) acetate pyruvate 2 -0.70 succinate → a-ketoglutarate 2 -0.67 acetate acetaldehyde 2 -0.60 3-phosphoglycerate → glyceraldehyde-3-P 2 -0.55 a-ketoglutarate isocitrate -0.38 NAD+ → NADH 2 -0.32 FMN-FMNH2...
Please do not answer if you are not sure, thank you Calculate E degree for the following half-reaction, FeS(s) + 2e^- Fe(s) + S^2- (aq) given that the solubility product for FeS at 25 degree C is 8.0 times 10^-19 and the standard reduction potential of the half-reaction shown below is -0.440 V. Fe2+ (aq) + 2e^- Fe(s) E degree_FeS = Construct the line notation of a cell that consists of a saturated calomel reference electrode, the FeS half-reaction shown...
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+...
In one step of glycolysis, glyceraldehyde 3-phosphate is oxidized by NAD+ to yield 3-phosphoglycerate and NADH. 1) Consider that reaction, in the direction written. Which chemicals) is(are) losing electrons, and which chemical(s) is(are) gaining electrons? It's best to answer this in a clear complete sentence. 2 2 2 Table 10-2 Standard Reduction Potentials for Redox Pairs of Biological Relevance* Redox Pair Number of (oxidized form → reduced form) Electrons E.(V) acetate pyruvate 2 -0.70 succinate → a-ketoglutarate 2 -0.67 acetate...
Please give an explanation as to why it is the correct answer, thank you. Use the following data to answer questions 13 - 16 €° (volts) Mn2+ + 2 e → Mn -1.03 Fe3+ + 3 e → Fe 9 -0.04 oglo biso 13. Which of the following statements is (are) true? a) The SHE would act as the cathode in a reaction with a Mn anode. b) Mn2+ will react spontaneously with Fe. c) Mn will reduce Fe to...
please answer all of the questions!! 1. Which substance is reduced in the following reaction (1 point): 2Al(s) + 3C12(g) → 2AlCl3(s) a. Al b. Cl2 c. AICI: d. None 2. Which substance is the reducing agent in the following reaction (1 point): 8H+ (aq) + MnO4 (aq) + 5Fe2+ (aq) → 5Fe3+ (aq) + Mn2+ (aq) + 4H2O(1) a. Fe2+ b. Mn04 c. H+ d. H20 3. Which substance is oxidized in the following reaction (1 point): Fes(s) +...
Use the table below as the basis for the calculations needed to answer the following questions. Standard Reduction Potentials for Redox Pairs of Biological Relevance Standard Redox Pair Number Reduction (oxidized form of reduced form) Potential, electrons Volts Acetate - pyruvate 2 -0.70 Succinate --- 2 -0.67 ketoglutarate Acetate 2 -0.60 acetaldehyde 3-phosphoglycerate glyceraldehyde-3 2 -0.55 IP arketoglutarate → 2 -0.38 isocitrate NAD+ -NADH 2 -0.32 FMN--FMNH2 2 -0.30 1,3- bisphosphoglycerate 2 -0.29 glyceraldehyde-3- IP Acetaldehyde → -0.20 ethanol Pyruvate-lactate...
Answer is NOT: -0.12 Part F What is the AE, for the reduction of lactate to pyruvate by NAD+ under standard conditions? Express your answer using two decimal places. Templates Symbols undo regio desde keyboard shortcuts help AEG' = V Submit Request Answer Use the table below as the basis for the calculations needed to answer the following questions. Standard Reduction Potentials for Redox Pairs of Biological Relevance Standard Redox Pair Number Reduction (oxidized form of reduced form) Potential, electrons...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
Write the half reactions and overall reaction for each cell with calculated overall potentials as shown in Table 5-1. (Note: for the iron solutions the Nernst equation must be used) Pb(s) | Pb(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) Cu(s) Zn(s) | Zn(NO3)2(0.1M) || Cu(NO3)2 (0.1M) Cu(s) Cds) | Ca(NO3)2 (0.1M) || Cu(NO3)2 (0.1M) | Cu(s) Cu() Cu(NO3)2(0.1M) Il Fe (0.1M/Fe? (0.1M graphite Pb(s) Pb(NO3)2(0.1M) Il Fe3(aq) (0.1M)/ Fe2(aq) (0.1MI graphite(s) Zns | Zn(NO3)2 (0.1M) || Pb(NO3)2 (0.1M) | Pb(s) Cdis Ca(NO3)2...