The free energy change for the following reaction at 25 °C, when [Cr³+] = 7.70x10-3 M and [Cu2+] = 1.12 M, is 66.7 kJ: Cr³+(7.70x10-3 M) + Cu+ (aq) Cr²+ (aq) + Cu²+(1.12 M) AG = 66.7 kJ What is the cell potential for the reaction as written under these conditions? Answer: →>> V Would this reaction be spontaneous in the forward or the reverse direction?

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.

Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr²+] = 0.768 M and [Fe2+] = 0.0170 M. Use the standard reduction potentials in this table. Cr(s) + Fe²+(aq) Cr² (aq) + Fe(s) E = x10 TOOLS V

K124.

Calculate the standard free‑energy change for this reaction at 25 °C. A list of standard reduction potentials can be found in this table.   2Au3+(aq)+3Ni(s)↽−−⇀2Au(s)+3Ni2+(aq) Δ?∘=   kJ/mol

What is the balanced reaction for the following galvanic cell?Cr(s)| Cr3+(aq) || Fe3+(aq) | Fe(s)   Fe(s) + Cr(s) ↔ Fe3+(aq) + Cr3+(aq)   Fe3+(aq) + Cr3+(aq) ↔ Fe(s) + Cr(s)   Fe3+(aq) + Cr(s) ↔ Fe(s) + Cr3+(aq)   Fe(s) + Cr3+(aq) ↔ Fe3+(aq) + Cr(s)

Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr2+]=0.876 M and [Ni2+]=0.0130 M Cr(s)+Ni2+(aq)↽−−⇀ Cr2+(aq)+Ni(s)

Determine the cell potential for the following cell Ni(s) | Ni2+(aq), 0.500 M || Ag+(aq), 0.010 M | Ag(s) The standard reduction potentials of Ni2+(aq) and Ag+(aq) are –0.25 V and +0.80 V, respectively.   A.  1.05 V   B.  1.10 V   C.  1.00 V   D.  1.16 V   E.  0.94 V

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+ (aq) + Hg(1)→ H₂(g) + Hg²+ (aq) 2H+ (aq) + 2e¯ → H₂(g) Fº red 2+ Hg2 (aq) + 2e¯ → Hg(1) AGO = || kJ K for this reaction would be Eº red = 0.000 V = = 0.855 V than one.