Nernst equation is `E = E^(@)-(RT)/(nF)lnQ`. If `Q = K_(C)`, then which one is one correct,

The Nernst equation , E = E^(0) -(RT)/(nF) . In Q indicates that the equilibrium constant K_(c) will be equal to Q when :

The Nernst equation E= E^(@) -RT/nF in Q indicates that the Q will be equal to equilibrium constant K_c when:

The Nernst equation E= E^(@) -RT/nF in Q indicates that the Q will be equal to equilibrium constant K_c when:

The Nernst equation E= E^(@) -RTlnQ in Q indicates that the Q will be equal to equilibrium constant K_c when:

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 equation, E^@ =(RT)/(nF)In K_c is called :

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 ?

for the electrorode reaction, M^(n+)(aq)+n e^(-)rarrM(s) Nernst equation is: E=E^o +frac(RT)(nF)ln frac(1)([M^(n+)] , E=E^o +RT ln[M^(n+)] , E=E^o +frac(RT)(nF)l n[M^(n+)] , E/E^o=frac(RT)(nF)l n[M^(n+)]. .