Question

3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO42−(aq) + 8 H2O(l) ΔG∘ = +87 kJ/molGiven the standard reduction potential of the half-reaction Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V, calculate the standard reduction potential of the half-reactionCrO42−(aq) + 4 H2O(l) + 3 e− → Cr(OH)3(s) + 5 OH−(aq)A. -0.43 VB. -0.28 VC. 0.02 VD. -0.13 VE. -0.15 V

Answer 1

The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V

Step-by-step explanation:

Step 1: 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

Step 2: 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

Step 3: 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

Step 4: 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