InterviewSolution
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InterviewSolution
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Calculate the emf of the cell. `Mg(s)|Mg^(2+) (0.2 M)||Ag^(+) (1xx10^(-3))|Ag` `E_(Ag^(+)//Ag)^(@)=+0.8` volt, `E_(Mg^(2+)//Mg)^(@)=-2.37` volt What will be the effect on emf If concentration of `Mg^(2+)` ion is decreased to `0.1 M` ? |
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Answer» Correct Answer - 3.0134`" V "`; increases Step I. Calculation of EMF of the cell Cell reaction : `Mg(s)+2Ag^(+)(1xx10^(-3)" M")to Mg^(2+)(0.2" M") +2Ag(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)])/([Ag^(+)]^(2))=[0.8-(-2.37)]-(0.0591)/(2)"log"(0.1)/((1xx10^(-4))^(2)` `=3.17-0.02955" log "(2xx10^(5))=3.17-0.02955xx5.3010` `=3.17-0.15664=3.0134" V "` Step II. Effect of decreasing concentration of `Mg^(2+)` ions on emf of cell The EMF of cell wil increase on decreasing the concentration of `Mg^(2+)` ions in solution. |
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