3ni2+(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

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890