An iron–sulfur protein in Complex III donates an electron to cytochrome c1. The reduction half-reactions and Ɛ°' values are shown below. Calculate the standard free energy change. Use 2 significant figures. FeS (ox) + e– → FeS (red) Ɛ°' = 0.280 V cyt c1 (Fe3+) + e– → cyt c1 (Fe2+) Ɛ°' = 0.215V ΔGo' = ? kJ · mol-1

Use standard reduction potentials to calculate the equilibrium constant for the reaction: Ni²*(aq) + 2Cr2*(aq)Ni(s) + 2Cr³*(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: AG° for this reaction would be than zero.

Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2+ Pb?*(aq) + 2F2²*(aq)→Pb(s) + 2FE³*(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: AG° for this reaction would be than zero.

Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energy and oxygen is the terminal electron acceptor. NO₂ + 6e-> NH4+ (+0.34 volts) • O₂ + 4e -> 2H₂O (+0.82 volts) ● Using the information given, calculate the AE for this reaction, balance the full reaction to determine the n, the number of electrons transferred when 30 moles of NH4* are oxidized. Finally, use the simplified Nernst Equation AG = -nFAE, where F = 96.5 kJ (mol e x V)-1 to determine the Gibbs Free energy available to do work! • Report your answer in kJ rounded to two decimal places. o Include trailing zeros!! Always report two decimal places even if the answer is a whole number o e.g. 18.00 not 18 • Report only the numeric portion of your answer o e.g. 1.01, not 1.01 kj per mole. • Answers should ALWAYS be negative since this is a spontaneous reaction. Type your answer...