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Calculate the e.m.f. of the following cell at 298K: 2Cr(s)+3Fe^(2+)(0.1M)to2Cr^(3+)(0.01M)+3Fe(s) Given: E_((Cr^(3+)//Cr))^(@)=-0.74V,E_((Fe^(2+)//Fe))^(@)=-0.44V.

Answer»


SOLUTION :`E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.44-(-0.74)=0.30V`
`E_(cell)=E_(cell)^(@)-(2.303RT)/(NF)"LOG"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))(n=6)`


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