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Calculate the standard cell potentials of galvanic cell in whiCHM the following reactions take place `:` `a. Cr(s) +3Cd^(2+)(aq) rarr 2Cr^(3+)(aq)+3Cd` `b. Fe^(2+)(aq)+Ag^(o+)(aq)rarr Fe^(3+)(aq)+Ag(s)` Calculate the `Delta_(r)G^(c-)` and equilibrium constant of the reactions . |
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Answer» Correct Answer - `a. Delta_(r)G^(c-)=-196.86 kJ mol^(-1),K_(c)=3.192xx10^(34)` `b. Delta_(r)G^(c-)=-2.895 kJmol^(-1),K_(c)=3.22` a. `E_("cell")^(c-)=E_("cathode")^(c-)-E_("anode")^(c-)` `=-0.40V-(-0.74)=+0.34V` `Delta_(r)G^(c-)=-nFE_("cell")^(c-)` `=-6 molxx96400C mol^(-1)xx0.34V` `=-196860J mol^(-1)=-196.86k J mol^(-1)` `Delta_(r)G^(c-)=--2.303RTlogK` `196860=2.303xx8.314xx298 log K` `logK=34.5014` `K="Antilog" (34.5014)=3.192xx10^(34)` `b`. `E^(c-)``_(cell)=+0.80V-0.77V=+0.03V` `Delta_(r)G^(c-)=-nFE_(cell)^(c-)` `=-(1 mol)xx(96500 C mol^(-1))xx(0.03V)` `=-2895J mol^(-1) Delta_(r)G^(c-)=-2.303RTlogK` `-2895=-2.303xx8.314xx298xxlogK` or log `K =0.5074` or `K ="Anitlog"(0.5074)=3.22` |
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