1.

Calcualte the emf of the following cell at 298 K: 2Cr(s) + 3Fe^(2+) (0.1 M ) to 2Cr^(3+) (0.01 M) + 3Fe(s) Given: E_("Cr^(3+)//Cr)^(@) = -0.74 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V.

Answer»

Solution :Standard emf of the CR - Fe cell is calculated as under:
`E_("cell")^(@) = E_("cathode")^(@) - E_("ANODE")^(@)`
`=-0.44 - (-0.74) = 0.30 V`
`E_("cell") = E_("cell")^(@) - 0.059/n log ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
Substituting the values, we have,
`E_("cell") = 0.30 - 0.059/6 log ([0.01]^(2))/([0.1]^(3)) = 0.30 -(-0.059//6) = 0.3098 V`


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