How is E^(0) cell related mathematically to the equilibrium constant K_(c) of the cell reaction ?
For a spontaneous reaction the DeltaG , equilibrium constant (K) and E_("cell")^(0) will be respectively .
In a reversible reaction, if the concentration of reactants are double, the equilibrium constant K will be x times the initial equilibrium constant. x will be equal to
In the equilibrium reaction, A(g) + 2B(g) + (g), the equilibrium constant, K_(c) is given by the expression
Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) Consider the two reactions: XeF_(6(g))+H_(2)O_((g)) harr XeOF_(4(g))+2HF_((g)), K_(1)" "XeO_(4(g))+XeF_(6(g)) harr XeOF_(4(g))+XeO_(3)F_(2(g)), K_(2) Then the equilibrium constant for the following reaction will be XeO_(4(g))+2HF_((g)) harr XeO_(3)F_(2(g))+H_(2)O_((g))
Passage: When all the coefficients in a balanced chemical equation are multiplied by a constant factor X the equilibrium constant (originally K) becomes K^J . Similarly, when balanced equations are added together, the equilibrium constant for the combined process is equal to the product of the equilibrium constants for each step. Equilibrium constant of the reversed reaction is numerically equal to the reciprocal of the equilibrium constant of the original equation. Unit of K_p = ("atm")^(Deltan) , Unit of K_c =("mol" L^(-1))^(Deltan) The equilibrium constants for the following reactions at 1400 K are given: 2H_2O_((g)) harr 2H_(2(g))+O_(2(g)) , K_1=2.1 xx 10^(-13) 2CO_(2(g)) harr 2CO_((g))+O_(2(g)) , K_2=1.4 xx 10^(-12) Then the equilibrium constant K for the reaction, H_(2(g))+CO_(2(g)) harr CO_((g)) + H_2O_((g)) is
The driving force DeltaG diminishes to zero on the way to equilibrium, just as in any other spontaneous process. Both DeltaG and the corresponding cell potential (E = (DeltaG)/(nF)) zero when the redox reaction comes to equilibrium. The Nernst equation for the redox process of the cell may be given as : E= E^(@) -(0.059)/( n) log Q The key to the relationship is the standard cell potential E^(@) , derived from the standard free energy change as : E^(@)=-(DeltaG^(@))/(nF) . At equilibrium, the Nernst equation is given as : E^(@) = -(0.059)/(n) log K At equilibrium , when K = 1 the correct relation is
The E^(@) values for the changes given below are measured against NHE at 27^(@) C Cu^(2+) + e to Cu^(+) , E^(@) = + 0.15 V , Cu^(+) + e to Cu, E^(@) = + 0.50 V , Zn^(2+) + 2e to Zn , E^(@) = - 0.76 V . The temperature coefficient of emf a cell designed as Zn|Zn^(2+) || Cu^(2+) || Cu is - 1.4 xx 10^(-4)v per degree . For a cell reaction in equilibrium DeltaG= 0 and DeltaG^(@) = -2.303 RT log K_(c) . The heat of reaction and entropy changes during the reaction are related by DeltaG = DeltaH - TDeltaS . The equilibrium constant constant for the reaction : Zn + Cu^(2+) iff Zn^(2+) + Cu is
Nernst equation gives the variation of potential of an electrode based on activity of ions temperature and pressure. The equation is E=E^(@) -(2.303RT)/(nF) logQ (or) E=E^@ - (0.0591)/(n) log Q E^@ = Standard potential and 'Q' is the reaction quotient. Which cell has least potential ?
