InterviewSolution
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InterviewSolution
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Calculate the e.m.f. of the cell, `Cr//Cr^(3+)(0.1 M) || Fe^(2+)(0.01 M)//Fe` `"Given" : E_(Cr^(3+)//Cr)^(@)=-0.75" V ", E_(Fe^(2+)//Fe)^(@)=-0.45" V "` `"Cell reaction" : 2Cr(s)+3Fe^(2+)(aq) to 2Cr^(3+)(aq)+3Fe(s)` `{"Hint". E_(cell)=E_(cell)^(@)-(0.0591V)/(6)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3)}` |
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Answer» Correct Answer - `+0.2606 V` Cell reaction : `2Cr(s)+3Fe^(2+)(0.01" M") to 2Cr^(3+)(0.1" M")+3Fe(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))=[(-0.45)-(-0.75)]-(0.0591)/(6)"log"((0.1)^(2))/((0.1)^(3))` `=0.30-(0.0591)/(6)log10^(4)=0.30-(0.0591xx4)/(6)=0.30-0.0394=+0.2606" V "` |
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