ChemistryThe EMF of the cell Zn | Zn2+ (0.01M) | | Fe2+ (0.001M) | Fe , at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction 10x. Find x.

The EMF of the cell Zn | Zn2+ (0.01M) | | Fe2+ (0.001M) | Fe , at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction 10x. Find x.

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    Solution:

    For this cell, reaction is;

    Zn+Fe2+Zn2++Fe 

    E=E°-0.0591nlogc1c2; E°=E+0.0591nlogc1c2

    E°=0.2905+0.05912log10-210-3=0.32 V

    At equilibrium,

    E°=0.05912log Keq

    log Keq=0.32×20.0591            =0.320.0295

    Keq=100.320.0295

    Comparing the value of 10x,

    x=0.320.0295=10.84710.85

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