Answer:
The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V
Explanation:
<u>Step 1</u>: Data given
3 Ni^2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO4^2−(aq) + 8 H2O(l) ΔG∘ = +87000 J/mol
Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V
<u>Step 2:</u> The half reactions:
Cathode: Ni2+(aq) + 2 e− → Ni(s) E° = -0.28 V
Anode: CrO4^2-(aq) + 4H2O(l) +3e- → Cr(OH)3(s) + 5OH- (aq) E°= unknown
<u>Step 3:</u> Calculate E°cell
ΔG° = -n*F*E°cell
⇒ with ΔG° = the gibbs free energy
⇒ n = the number of electrons in the net reaction = 6
⇒ F = the Faraday constant = 96485 C
⇒ E°cell= the standard cell potential
<u>Step 4:</u> Calculate E°(Cr6+/Cr3+
E°cell= ΔG°/(-n*F)
E°cell = 87000 /(-6*96485)
E°cell = -0.150 V
E°cell = E°(Ni2+/Ni) - E°(Cr6+/Cr3+)
E°(Cr6+/Cr3+) = -0.13V
The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V
