The EMF of the following cell is found to be 0.20 V at 298 K `Cd|Cd^(2+)(?)||Ni^(2+)(2.0M)|Ni` What is the molar concentration of `Cd^(2+)` ions in the solution? `(E_(Cd^(2+)//Cd)^(@)=-0.40V,E_(Ni^(2+)//Ni)^(@)=-0.25V)`

The EMF of the following cell is found to be 0.20 V at 298 K `Cd|Cd^(2

1.	The EMF of the following cell is found to be 0.20 V at 298 K `Cd\|Cd^(2+)(?)\|\|Ni^(2+)(2.0M)\|Ni` What is the molar concentration of `Cd^(2+)` ions in the solution? `(E_(Cd^(2+)//Cd)^(@)=-0.40V,E_(Ni^(2+)//Ni)^(@)=-0.25V)`
Answer» The cell reaction is: `Cd+Ni^(2+)toCd^(2+)+Ni" "E_(Cell)^(@)=-0.25-(-0.40)==0.15V` Applying nernst equation, `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cd^(2+)])/([Ni^(2+)])` `0.20=0.15-(0.0591)/(2)"log"([Cd^(2+)])/(2)" or "log[Cd^(2+)]-log2=-1.690` or `log[Cd^(2+)]=-1.690+0.3021=-1.3879" or "[Cd^(2+)]="antilog "overline(2).6121=0.0409`M

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The EMF of the following cell is found to be 0.20 V at 298 K `Cd|Cd^(2+)(?)||Ni^(2+)(2.0M)|Ni` What is the molar concentration of `Cd^(2+)` ions in the solution? `(E_(Cd^(2+)//Cd)^(@)=-0.40V,E_(Ni^(2+)//Ni)^(@)=-0.25V)`

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