Question

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 -​

Answer 1

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