Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Cu²+ (aq) + Fe(s)→ Cu(s) + Fe²+ (aq) Cu2+ (aq) + 2e Fe2+ (aq) + 2e Submit Answer → Cu(s) Ee Cu(s) = 0.337 V red AG: kj K for this reaction would be → Fe(s) E Fe(s) E= -0.440 V red e than one. Retry Entire Group 9 more group attempts remaining

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Pb²+ (aq) + Cu(s) - → Pb(s) + Cu²+ (aq) AG° Pb²+ (aq) + 2e¯¯ → Pb(s) Ee Cu²+ (aq) + 2e¯ → Cu(s) Fre red = kJ K for this reaction would be = -0.126 V = 0.337 V + than one.

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2Cu²+ u²+ (aq) + 21- (aq) → 2Cut (aq) + 12 (8) 2+ Cu²+ (aq) + e → Cut (aq) Ee red I2 (s) +2e → 21¯ (aq) AGO = K for this reaction would be kJ E red - 0.153 V 0.535 V than one.

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.