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3.4     Calculate the standard cell potentials of galvanic cell in which the following reactions take place:

          Fe^{2+}(aq)+Ag^{+}(aq)\rightarrow Fe^{3+}(aq)+Ag(s)

          Calculate the \Delta _{r}G^{e}and equilibrium constant of the reactions.

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The galvanic cell of the given reaction is shown below :-

                                              

We can know about the electrode potential of Fe and Ag with the help of table of standard electrode potential.

We have :                          E_{(cell)}^{\circ} = E_{R}^{\circ} - E_{L}^{\circ}

or                                                     = 0.80 - 0.77

or                                                     = 0.03\ V

Now consider :                    \Delta G_r^{\circ} = -\ nFE_{(cell)}^{\circ}

or                                                     = -1\times96487\times0.03

                                                        = -2.89\ KJ\ mol^{-1}

Now for equilibrium constant :

                                             log\ K =\ -\frac{\Delta G_r^{\circ}}{2.303\times RT}

or                                                        =\ -\frac{-2894.61}{2.303\times 8.314\times298}

or                                                         =\ 0.5073

Thus                                              k\ \approx \ 3.2

Posted by

Devendra Khairwa

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