Answer to Question #199981 in Chemistry for Julius Sarmiento

Question #199981

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


1
Expert's answer
2021-05-31T02:25:11-0400

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)


cell = E°cathode - E°anode

cell = 0.77 V + 0.74 V = 1.51 V

cell = 1.51 V


Answer: E°cell = 1.51 V

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