Saved Bookmarks
| 1. |
Calculate the standard cell potentials of galvanic cell in which the following reactions take place: (i) 2Cr (s) + 3Cd^(2+) (aq) to2Cr^(3+) (aq) + 3Cd (ii) Fe^(2+)(aq) + Ag^(+)(aq) to Fe^(3+)(aq) +Ag (s) Given: E_(Cr^(3+),Cr) = -0.74 V, E_(Cd^(2+),Cd)^(@) = -0.04 V, E_(Ag^(+),Ag) = 0.80 V, E_(Fe^(3+),Fe^(2+))^(@) = 0.77 V. Calculate the Delta_(r)G^(@) and equilibrium constant of the reactions. |
|
Answer» Solution :USING the following relations and substituting the values, we get (i) `E_(cell)^(@) = E_("cathode")^(@) -E_("ANODE")^(@) = -0.40 V -(-0.74 V) = +0.34` V `Delta_(R)G^(@) =-nFe_("cell")^(@) =-6 mol xx 96500 C mol^(-1) = -196.86 kJ mol^(-1)` `=-19600 C V mol^(-1) = -19860 J mol^(-1) = -196.86 kJ mol^(-1)` `-Delta_(r)G^(@) =2.303 RT log K` or 196860 = `2.303 xx 8.314 xx 298 log K` or log K = 34.5014 (ii) `E_("cell")^(@) = +0.80 V - 0.77 v = 0.03 V` `Delta_(r)G^(@) =-nFE_("cell")^(@) =-(1 mol ) xx (96500C mol^(-1)) xx (0.03 V)` `K = "Antilog" (0.5074) = 3.22` |
|