The EMF of the cell
Cr 1 Cr+3 (0.1M) || Fe+2 (0.01M) I Fe
(Given EC-+3/C+ = -0.75V, Efeta; Fe = -0.45 V)
Half cell reactions are -
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Answer:
0.26053 V
Explanation:
anode oxidation reaction :
2Cr – 6e– > 2Cr3+
cathode reduction reaction :
3 Fe2+ + 6e – > 3Fe
complete reaction:
2Cr + 3 Fe2+ > 2Cr3+ + 3Fe
E°cell
=[E°( Cr/ Cr3+) + E°(Fe2+/Fe)] –(0.0592/n)log{[Cr3+]^2 / [Fe2+]^3}
=[0.75–0.45]–(0.0592/6)log[(0.1)^2/(0.01)^3]
=0.26053 V
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