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Calculate the value of Ecell at 298K for the following cell:
Al/Al3+(0.01M)||Sn2+(0.015M)/Sn
EAl3+/Al=1.66V and ESn2+/Sn=0.14V

Answer
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Hint:Here we are given with the Ecell of the reaction but we have to find the Ecell. So for the relation between Ecell and Ecell we have an equation which is known as Nernst equation. First we will find the value of Ecell from the given data and then we have to apply Nernst equation to find Ecell of the reaction.

Complete step by step answer:
First of all let us talk about how to find Ecell of the reaction when you know the values of E of its components i.e. the ions or atoms which are reduced or oxidised in the reaction.
Reduced ions or atoms are that in which there is decrease in the positive charge on the ion i.e. gain of electrons.
Oxidised ions or atoms are that in which there is an increase in the positive charge on the ion i.e. loss of electrons.
Now if in the reaction we know which is oxidised and which is reduced atom and is we also know the values of E of its components i.e. the ions or atoms. Then we can calculate the value of Ecell of the reaction as: Ecell=EREL where ER is the E of the atom which is at right side and EL is the E of the atom which is at left side. Here in the question the atom on the right side is Sn and the atom on left side is Al. So the Ecell of the reaction will be Ecell=EREL, where ER=0.14V and El=1.66V. So Ecell=0.14(1.66)=1.52V.
And if we try to write down net cell reaction then it will be as:
2Al+3Sn2+2Al3++3Sn.
Here in the reaction we have total 6 transfer electrons.
Nernst equation: It is defined as the relation of Ecell and Ecell which is as:
Ecell=Ecell0.059nlog[Aq+]p[Bp+]q if the equation is as pA+qBp+pAq++qB where [Aq+] is the concentration of the ion which is oxidised and [Bp+] is the concentration of the ion which is reduced in the reaction. And p is the coefficient of the compound which is oxidised and q is the coefficient of the compound which is reduced in the reaction and n is the total number of electrons transferred.
Here in the question we are given with the reaction as 2Al+3Sn2+2Al3++3Sn. Here we have A=Al,B=Sn,p=2,q=3,[Al3+]=0.01M,[Sn2+]=0.015M and the value of n is 6
Substituting these values in the Nernst equation we get:
Ecell=0.1520.059nlog[0.01]2[0.015]3
Then if we solve further after taking log of this then log of the required term is 1.47.

So the Ecell=1.520.01447=1.50553V.

Note:
Ecell of a reaction is defined as electrode potential of the cell and Ecell of a reaction is defined as electrode potential measured at 1 atmosphere pressure, 1 molar solution at 25C also known as standard electrode potential.
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