Given the following standard reduction potentials: MnO4 (aq) + 8H*(aq) + 5 e - - Mn2*(aq) + 4 H20() E° = 1.51 V Fe3+ (aq) + e Fe<* (aq) E° = 0.77 V What is the cell potential E for the following reaction under the non-standard condition with 0.16 M of MnO4", 0.92 M of Mn2*, 0.96 M of Fe+, and 0.19 M of Fe2* in a buffered solution with 1.0 M of H* ? Please keep your answer to two decimal places. 5FE2* (aq) + MnO4 (aq) + 8H*(aq) (aq) + Mn2+ (aq) + 4H20 () --> 5FE3+

Ff.39.

3.) Consider the following standard reduction potentials, which statement is correct at standard conditions? Ni?* (aq) + 2 e → Ni (s) E°red = -0.26 V I2 (s) + 2 e -> 2 (aq) E°red = 0.54 V I. Ni²* (aq) is a stronger oxidizing agent than I2 (s), and I is a stronger reducing agent than Ni (s) II. I2 (s) is a stronger oxidizing agent than Ni2", and Ni (s) is a stronger reducing agent than I (aq) III. Ni (s) is a stronger oxidizing agent than I (aq), and Ni2* is a stronger reducing agent than I2 (s) IV. I(aq) is a stronger oxidizing agent than Ni (s), and I2 (s) is a stronger reducing agent than Ni2+ (aq) А. IV В. Ш С. П D. I

Consider the following standard reduction potentials: F2(g) + 2 e - 2 F (aq) E° = +2.87 V Mg2*(aq) + 2 e - Mg(s) E° = -2.37 V Under standard conditions, O Mg(s) is a stronger oxidizing agent than F (aq), and Mg2*(aq) is a stronger reducing agent than F2(g), O F (aq) is a stronger oxidizing agent than Mg(s), and F2(g) is a stronger reducing agent than Mg2*(aq). O F2(g) is a stronger oxidizing agent than Mg-"(aq), and Mg(s) is a stronger reducing agent than F (aq). O Mg2*(aq) is a stronger oxidizing agent than F2(g), and F(aq) is a stronger reducing agent than Mg(s).

Use the standard reduction potentials located in the "Tables' linked above to calculate the standard free energy change in kJ for the reaction: 3Hg"(aq) + 2Cr(s)3Hg(1) + 2Cr*(aq) Answer: |kJ K for this reaction would be than one.

1) Calculate the following for the reaction at 298.0 K . The reaction is 2 SmCl3(aq) + 6 LiCl(aq) + 6 e- → 2 Sm(s) + 6 Li+(aq) + 6 Cl-(aq) A) The standard Gibbs free energy change of the reaction @ 298.0 K B) The equilibrium constant for the reaction C) The standard cell potential 2) Calculate the free energy change and the cell potential for the reaction at 298.0 K if all reactants are at 2.00 M and/or 2.00 atm and all products are at 0.0100 M and/or 0.0100 atm

Examine the following half-reactions and select the strongest oxidizing agent among the substances. v Jersey Departm E° = 0.755 V [PtCl4]²¯(aq) + 2e¯ 2 Pt(s) + 4CI'(aq); RuO4(s) + 8H*(aq) + 8e¯ 2 Ru(s) + 4H2O(1); FeO4²(aq) + 8H*(aq) + 3e¯ 2 Fe3*(aq) + 4H2O(/); H4XEO6(aq) + 2H*(aq) + 2e¯ 2 XeO3(aq) + 3H2O(/); E° = 2.42 V E° = 1.038 V E° = 2.07 V RuO4(s) H4XEO6(aq) [PtCl4]² (aq) O HFEO4¯ (aq) CI(aq)

Standard Reduction Potentials (Volts) at 25 °C F2 (g) + 2 e-  2 F- (aq) 2.870 O2(g) + 4 H3O+(aq) + 4 e-  6 H2O(l)  1.229 2 H2O(l) + 2 e-  H2(g) + 2 OH-(aq) -0.828 Fe2+ (aq) + 2 e-  Fe (s) -0.440   A 0.122 M neutral aqueous FeF2 solution is electrolyzed under 1 atm pressure using platinum electrodes.(a) Write the half-reactions expected to occur, taking into account the effect of the non-standard conditions on the electrode potentials.Half-reaction at anode:     +       +       ->       +       +     Half-reaction at cathode:     +       +       ->       +       +     (b) What is the expected decomposition potential? _______ V