Question

Given:

Ag+(aq) + e-→Ag(s)    E° = +0.799 V

Cr3+(aq) + e-Cr2+(aq.) Eo= - 0.41 V

Ni2+(aq) + 2 e-→Ni(s) E° = -0.267 V

Generate two voltaic cells using the above 3 reactions. Show the anode and cathode half-reactions and the overall cell reactions and calculate Eo cell values.

Answer 1

At cathode high reduction value equation will occur

Ag+ + e = Ag (s). E° = 0.799 V

At anode low reduction value will occur

Cr3+(aq) = Cr2+(aq) + e. E° = 0.41 V

Net cell reaction

Ag+(aq) + Cr3+(aq) = Ag(s) + Cr2+(aq) E° = 0.799 + 0.41 = 1.209 V

2. Ni2+(aq) + 2e = Ni(s) E° = -0.267 V

2Cr2+(aq) = 2 Cr3+(aq) + 2e E°= 0.41 V

Net cell reaction

Ni2+(aq) + 2 Cr2+(aq) = 2Ni(s) + 2 Cr3+(aq)

E° = 0.41 + ( -0.267 )

= 0.143 V

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