Calculate the gibbs free energy (ΔGo)  in kilojoules (KJ) for ColCo2+ II Cd2+I Cd   The faraday constant is 96495 couloumbs/mole of electron.   Half Reactions for the cell: Co2+(aq) + 2 e–→ Co(s)  Eo  = –0.277 V Cd2+(aq) + 2 e–→ Cd(s) Eo = –0.403 V Group of answer choices 24.3 KJ -24.3 KJ 12.2 KJ -12.2 KJ

An electrochemical cell is based on these two half-reactions: Oxidation: Cu(s) → Cu2+(aq, 0.010 M) + 2 e− Reduction: MnO4 −(aq, 2.0 M) + 4 H+(aq, 1.0 M) + 3 e− → MnO2(s) + 2 H2O(l) Calculate the cell potential.

7. Please answer the following questions. (a) Given the following reduction potentials Cd (OH)2 (s) + 2e → Cd (s) + 2 OH (aq) Eºred = -0.76 V NIO(OH) (s) + H₂O (1) + e → Ni(OH)2 (aq) + OH (aq) Eºred = +0.49 V Calculate the standard emf of the nickel-cadmium (nicad) rechargeable battery. *hint: the reaction for the discharge of nicad battery is: Cd (s) + 2 NIO(OH) (s) + 2 H₂O (1)→ 2Ni(OH)2 (aq) + Cd(OH)2 (s) (b) A typical nicad voltaic cell generates an emf of +1.30 V. Why is there a difference between this value and the one you calculated in part (b)? (c) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

Calculate the cell potential, E, for the given reactions at 25.00 "C using the ion concentrations provided. Then, determine if th cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E). Pt(s) + Fe²(aq) = P² (aq) + Fe(s) [Fe²*1 = 0.0064 M E= incorrect -4761 E= Cu(s) + 2 Ag* (aq) = Cu² (aq) + 2 Ag(s) |Cu²+1 = 0.027 M Ag1=0.027 M E [P1²*1 = 0.067 M Co² (aq) + Ti (aq) — Co¹ (aq) + Ti² '(aq) Co²1 = 0.055 M Co= 0.030 M ITP 1 = 0.0070 M T² = 0.0107 M mam V The cell is not spontaneous spontaneous. The cell is spontaneous not spontaneous. incorrect The cell is spontaneous not spontaneoux.