If the value of equilibrium constant (K_f) for the operation Zn_((aq))^(2+) + 4OH_((aq))^(-) iff Zn(OH)_(4(aq))^(2-) is 10^(10x), then x is (Given : Zn_((aq))^(2+) + 2e^(-) to Zn_((s)) , E^(@)=-0.76 V, Zn(OH)_(4(aq))^(2-) + 2e^(-) to Zn_((s)) + 4OH_((aq))^(-) , E^(@) = - 1.36 V, 2.303 (RT)/(F) = 0.06)

If the value of equilibrium constant (K_f) for the operation Zn_((aq))

1.	If the value of equilibrium constant (K_f) for the operation Zn_((aq))^(2+) + 4OH_((aq))^(-) iff Zn(OH)_(4(aq))^(2-) is 10^(10x), then x is (Given : Zn_((aq))^(2+) + 2e^(-) to Zn_((s)) , E^(@)=-0.76 V, Zn(OH)_(4(aq))^(2-) + 2e^(-) to Zn_((s)) + 4OH_((aq))^(-) , E^(@) = - 1.36 V, 2.303 (RT)/(F) = 0.06)
Answer» Solution :(1) `ZN^(+2) + 2e^(-) to Zn ,(-0.76 xx 2)` (2) `Zn + 4OH^(Ɵ) to [Zn(OH)_4]^(-2) + 2e^(-) , ( + 1.36 xx 2)` (3) `Zn^(+2) + 4OH^(Ɵ) to [Zn(OH)_4],(E_(3)^(0) xx 2 ) (3) = (1) + (2)` ` E_(3)^(0) = ((-0.76 xx 2)+ (1.36 xx 2))/(2) ,E_(3)^(0) = + 0.60 V = (0.0591)/(2) log K_c` `log K_(c) = (0.60 xx 2)/(0.0600 ) =2 xx 10^(2) , log K_(c) = 10 xx 2 = 20 , K_(c) (10)^(10 xx 2) , K_(c)(10)^(10xx2) = (10)^(10 x) , x = 2 `