Calculate
Interpretation:
The equilibrium constant
Concept introduction:
The standard cell potential of a particular cell is given as:
Here,
The relation between the cell potential
The relation between standard Gibbs free energy
Answer to Problem 24QP
Solution:
(a)
(b)
(c)
Explanation of Solution
a)
This reaction can be written astwo half-reactions, as follows:
Fromtable
Now, the standard cell potential of the cell is given as:
Substitute the values of half-cell potential in the above expression.
The relation between standard Gibbs free energy
Here, the number of exchange of electrons is two;
Thus, the value of
Substitute the values of Faraday constant, the number of electrons transferred in the half-cell and the standard cell potential.
The relation between the cell potential
Substitute the values
Thus, the equilibrium constant of this reaction is
b)
This reaction can be written astwo half-reactions, as follows:
Fromtable
The standard cell potential of a cell is given as follows:
Substitute the values of half-cell potential in the above expression:
The relation between standard Gibbs free energy
Here, the number of exchange of electrons is four;
Thus, the value of
The Gibb’s energy is calculated by the expression as follows:
The relation between the cell potential
Substitute the values of
Thus, the equilibrium constant of this reaction is
c)
This reaction can be written astwo half-reactions as follows:
Fromtable
The standard cell potential of a cell is given as follows:
Substitute the values of half-cell potential in the above expression,
The relation between standard Gibbs free energy
Here, number of exchange of electrons is six;
Thus, the value of
Substitute the values of
The relation between the cell potential
Substitute the values of
Thus, the equilibrium constant of this reaction is
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Chapter 19 Solutions
EBK CHEMISTRY
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