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

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

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