1.

Find out work done for the given cell Cr|Cr^(+3) || Fe^(+2)| Fe E_(Cr//Cr^(3+))^(@) = 0.74 V, E_(Fe^(2+)//Fe)^(@) = -0.44 V

Answer»

Solution :`E_("cell")^(@) = E_("cathode")^(@) - E_("ANODE")^(@)`
`=-0.44 V + 0.74 V = 0.30` V
[since cell reaction `2Cr + 3Fe^(2+) to 2Cr^(3+) + 3Fe`, n = 6 mol ELECTRON]
`W= -nFE^(@) =-6 xx 96500 xx 0.30 W = 173.7` KJ


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