A voltaic cell is set up at `25^(@)C`with the following half cells : `Al^(3+)`(0.001 M) and `Ni^(2+)`(0.50 M) Write the equation for the reaction when the cell generates the electric current. Also determine the cell potential (Given `E_(Ni^(2+)//Ni)^(@)=-0.25 V, E_(Al^(3+)//Al)^(@)=-1.66 V)`.

A voltaic cell is set up at `25^(@)C`with the following half cells : `

1.	A voltaic cell is set up at `25^(@)C`with the following half cells : `Al^(3+)`(0.001 M) and `Ni^(2+)`(0.50 M) Write the equation for the reaction when the cell generates the electric current. Also determine the cell potential (Given `E_(Ni^(2+)//Ni)^(@)=-0.25 V, E_(Al^(3+)//Al)^(@)=-1.66 V)`.
Answer» Correct Answer - 1.4602V The cell reaction is : `2Al(s)to 2Al^(3+)(aq)+6e^(-)` `(3Ni^(2+)(aq)+6e^(-)to3Ni(s)" ")/(2Al(s)+3Ni^(2+)(aq)to Al^(3+)(aq)+3Ni(s))` `E_(cell)^(@)=E_(Ni^(2)(aq)//Ni(s))^(@)-E_(Al^(3+)(aq)//Al(s))^(@)=(-0.25) -(-1.66)=-0.25+1.66=+1.14"V "` According to Nernst equation : `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Al^(3+)]^(2))/([Ni^(2+)]^(3))=1.41-(0.0591)/(6)"log"((0.001)^(2))/((0.5)^(3))` `=1.41-(0.0591)/(6)((10^(-6)xx10^(3)))/(125)=1.41-(0.0591)/(6)"log"(log8xx10^(-6))` `=1.41-(0.0591)/(6)[log8-6log10]=1.41-(0.0591)/(6)(0.9031-6)` `=1.41-(0.0591)/(6)(-5.0969)=1.41-(0.3012)/(6)=1.41+0.0502=1.4602" V "`

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A voltaic cell is set up at `25^(@)C`with the following half cells : `Al^(3+)`(0.001 M) and `Ni^(2+)`(0.50 M) Write the equation for the reaction when the cell generates the electric current. Also determine the cell potential (Given `E_(Ni^(2+)//Ni)^(@)=-0.25 V, E_(Al^(3+)//Al)^(@)=-1.66 V)`.

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