Answer to Question #199981 in Chemistry for Julius Sarmiento

Question

Calculate the cell potential of Voltaic cell. Cr3+ + 3e - Cr E0 = -0.74 V Fe3+ + e - Fe2+ E0 = +0.77 V Cr + 3Fe3+ - 3Fe2+

Answer 1

Solution:

Cr³⁺ + 3e- ⟶ Cr, E° = -0.74 V

Fe³⁺ + e- ⟶ Fe²⁺, E° = +0.77 V

Cathode (reduction): Fe³⁺(aq) + e- ⟶ Fe²⁺(aq), E°_red = 0.77 V

Anode (oxidation): Cr(s) ⟶ Cr³⁺(aq) + 3e-, E°_ox = -(-0.74) V

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Cr(s) + 3Fe³⁺(aq) ⟶ 3Fe²⁺(aq) + Cr³⁺(aq)

E°_cell = E°_cathode - E°_anode

E°_cell = 0.77 V + 0.74 V = 1.51 V

E°_cell = 1.51 V

Answer: E°_cell = 1.51 V