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Nernst Equation Calculator
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Nernst Equation may is something which calculates the electromagnetic force of the cells when concentration of reactants and products of the cell reactions are known.
Nernst Equation Calculator calculates the Nernst equation at room temperature Eo if Standard Half cell reduction potential Eo, z = No of moles transferred to the cell, Chemical Activity for the reduction species ared and Chemical Activity for the oxidation species ared are given.
 

Steps for Nernst Equation Calculator

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Step 1 :  

Read the problem and list out the given quantity.



Step 2 :  

Use the formula:


E = Eo - $\frac{0.05916}{n}$ log10 $\frac{a_{Red}}{a_{Ox}}$


Where Eo = Standard Half cell reduction potential


z = No of moles transferred to the cell


ared =Chemical Activity for the reduction species


aox = Chemical Activity for the oxidation species


Substituting the values, we get the answer.



Problems on Nernst Equation Calculator

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  1. The Emf of the following cells are:

    Mg(S)| Mg2+ (0.2 M) | | Ag+ (1 $\times$ 10-3)| Ag

    Find the Reduction potential for this.


    Step 1 :  

    given : Eo Ag+/Ag = 1 V


                Eo Mg2+/Mg = - 3V


    Eo = Eo Ag+/Ag + Eo Mg2+/Mg = 1 + 3 = 4 V


    Chemical Activity for the reduction species (ared) = 0.2


    Chemical Activity for the oxidation species (aoxi) = 0.001.


    Number of Moles of Electrons Transferred in the Cell (z) = 2.



    Step 2 :  

    The Cell reaction is Mg + 2Ag+  --> 2Ag + Mg2+.


    E = Eo - $\frac{0.05916}{z}$ log10 $\frac{a_{Red}}{a_{Ox}}$


       = 4 - $\frac{0.5916}{2}$ log10 $\frac{0.2}{0.001}$


      = 4 - 0.02958 $\times$ 2.301.


      = 4 - 0.068


       = 3.932.



    Answer  :  

    The Reduction Potential Eo = 3.932 V.



  2. The Emf of the following cells are:

    Zn| Zn2+ (1 M) | | Ag+ (1 M)| Ag

    Find the Reduction potential for this if EoAg+/Ag= - 0.5 V and EoMg2+/Mg = + 1V.


    Step 1 :  

    given : Eo Ag+/Ag = - 0.5 V


                Eo Mg2+/Mg = + 1V


    Eo = Eo Ag+/Ag + Eo Mg2+/Mg = 0.5 + 1 = 1.5 V


    Chemical Activity for the reduction species (ared) = 1


    Chemical Activity for the oxidation species (aoxi) = 1.



    Step 2 :  

    Reduction Potential is


    E = Eo - $\frac{0.05916}{z}$ log10 $\frac{a_{Red}}{a_{Ox}}$


       = 1.5 - $\frac{0.5916}{2}$ log10 $\frac{1}{1}$


      = 1.5 - 0.02958 $\times$ 0.


      = 1.5 - 0.


       = 1.5 V.



    Answer  :  

    The Reduction Potential Eo = 1.50 V.



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