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![We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2
Solution
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2](https://image.slidesharecdn.com/weknowthatfromnernstequationeocelleo0-230412101722-b2cf419f/75/We-know-that-from-Nernst-Equation-E-o-cell-E-o-0-059-n-pdf-1-2048.jpg)
The document discusses the application of the Nernst equation to calculate the cell potential (E°cell) in an electrochemical reaction. It provides specific values for standard potential (E°) and concentrations for products and reactants, including cadmium ions (Cd²⁺) and nickel ions (Ni²⁺). The document highlights that two electrons are transferred in the reaction.
![We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2
Solution
We know that from Nernst Equation ,
E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ])
Given Eo cell = 0.17 V
Eo = 0.24 V
[products ] = [Cd 2+ ] = ?
[reactants ] = [ Ni 2+ ] = 1.0 M
N = no . of electrons transferred = 2](https://image.slidesharecdn.com/weknowthatfromnernstequationeocelleo0-230412101722-b2cf419f/75/We-know-that-from-Nernst-Equation-E-o-cell-E-o-0-059-n-pdf-1-2048.jpg)