Write the Nernst equation for the following half-reaction and find E when pH =3.00 and PASH, = 1.00 mbar. As(s)+3H +3e ASH,(g); E° = -0.238 V %3D %3D (A) -0.356 V (B) +0.556 V (C) +0.356 V (D) -0.556 V

Given the following standard electrode potentals: Ag°(aq) + e = Ag(s) E° = 0.80 V N,(g) + 5H¨(aq) + 4e¯ = N,H;(aq) E° =-0.23 V Fe* (aq) + 2e = Fe(s) E° =-0.44 V Which of the reactions is spontaneous? N,(g) + 2Fe(s) + 5H¨(aq) = N,H (aq) + 2Fe²* N,(g) + 4Ag(s) + 5H¨(aq) = N‚H; (aq) + 4Ag* Fe* + 2Ag(s) 2 2Ag*(aq) + Fe(s) None of the above are spontaneous.

agent? (a) Ag (aq) 20. Which statement is true considering the following reactions? Sn² (aq) +2e-> Sn (s) E = -0.136 V Fe" (aq) +3e > Fe (s) E = -0.036 V (a) The tin electrode is the cathode and electrons flow from anode to cathode (b) The tin electrode is the cathode and electrons flow from cathode to anode (c) The tin electrode is the anode and electrons flow from anode to cathode (d) The tin electrode is the anode and electrons flow from cathode to anode 21 Which of the following r reser (b) Ag (s) (c) Zn² (aq) (d) Zn (s)

3. (a) b C. d. e. f. Given 0.01 Molar concentration Au³+ ion and 0.01 molar of Ag+ ion, and the following reaction: or Au³(aq) Ag+ (aq) -→ Au (s) -→ Ag (s) Write the balanced oxidation reduction equation for the cell E° = 1.52 V Eº = 0.80 V Which specie is being oxidized Which Specie is at the anode? Which Specie is the oxidizing agent? Which specie lost electron/s? Which specie saw an increase in oxidation state? Calculate the Ecell at non-standard condition for the reaction

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Given the following half-reactions and their standard reduction potentials, 2CIO,- + 12H+ + 10e Cl, + 6H2O Eao,- = 1.47 V S,0,2- + 2e= 2S0,2- Eo = 2.01 V calculate the following: (a) E° cell for the reaction: E° cell (b) AG° for the cell reaction: x 10 kJ (c) the value of K. for the cell reaction: x 10

Please don't provide hand writtin solution....

R = 8.314 mol·K F = 96,485 mot AG = AG° + RT · In(Q) ΔΕΔΕ-).In(O) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (1) 1.229 2 z" (aq) Z2 (s) + 2 e 3+ (aq) + 3 е 0.426 A (s) 0.292 2 H20 (1) + 2 e G2+ (aq) + 2 e H2 (g) + 2 OH (aq) - 0.828 G (s) - 1.245 > M2+ (aq) + 2 e - 1.893 → M (s) A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. Consider the same cell from the above prompt. Calculate AG° in kJ/mol for this galvanic cell. Report your answer with 4 significant figures. You do not need to report units with your answer. If your value is negative, make sure to include a "-" symbol.

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)

I need the solution (with work) for D