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  • For the reaction <img alt="Mg + 2Ag^{+}\rightarrow Mg^{2+}+ 2A g" src="https://entrancecorner.oncodecogs.com/gif.latex?Mg%20&plus;%202Ag%5E%7B&plus;%7D%5Crightarrow%20Mg%5E%7B2&plu

For the reaction Mg + 2Ag^{+}\rightarrow Mg^{2+}+ 2A g

E^0 = 3.17 V , find the value of  \Delta G^0(in kJ)

Option: 1

-612


Option: 2

-603


Option: 3

-598


Option: 4

-591


Answers (1)

As we have learnt,

Feasibility and Gibbs Free Energy of Reaction -

Let n faraday charge be taken out of a cell of emf E, then work done by the cell will be calculated as:
Work = Charge × Potential
Work done by the cell is equal to the decrease in the free energy.

-\Delta \mathrm{G}=\mathrm{nFE}

Similarly, maximum obtainable work from the cell at standard condition will be:

\mathrm{W}_{\max }=\mathrm{nF} \mathrm{E}_{\mathrm{cell}}^{0} \quad \text { where } \mathrm{E}_{\mathrm{cell}}^{0}=\text { standard emf of standard cell potential }

\mathrm{-\Delta G^{\circ}=n F E_{\text {cell }}^{0}}

-

\Delta G^0= -nFE^0

= -2 * 96487*3.17 =-611.7kj\approx -612kj

 

Posted by

Sumit Saini

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