The standard Gibbs energy for the given cell reaction in `kJ mol^(-1)` at 298 K is : `Zn(s)+Cu^(2+)(aq)rarrZn^(2+)(aq)+Cu(s)`, `E^(@)=2 V` at `298 K` (Faraday's constant, `F = 96000 C mol^(-1)`)
The standard Gibbs energy for the given cell reaction is KJmol^(-1) at 298 K is : Zn(s)+CU^(2+) (aq) rightarrow Zn^(2+) (aq) + Cu(s) E^(@)= 2V at 298 K (Friday's constant , F = 96000 C mol^-1 )
The standard Gibbs energy for the given cell reaction is -3.84 xx 10^(x) J/mole at 298 K. The numerical value of x is______. Zn(s)+Cu^(2+)(aq) rarr Zn^(2+)(aq)+Cu(s) E^(@)=2V at 298 K (Faraday’s constant, F = 96000 C "mol"^(-1) )
What is the Gibbs energy of the following reaction? Zn(s)+Cu^(2+) (aq) rarr Zn^(2+)(aq)+Cu^(2+) (s), E_("cell")^(@)=1.1 V
Calculate DeltaG^(@) and log K_(c) for the following reaction at 298 K : 2Al(s)+3Cu^(2+)(aq)rarr2Al^(3+)(aq)+3Cu(s) Given E_(cell)^(@) =2.02 V
Write Nernst equation for the following cell reaction : Zn(s) | Zn^(2+)(aq)|| Cu^(2+)(aq) | Cu(s)
Calculate the standard free enegry change for the reaction: Zn + Cu^(2+)(aq) rarr Cu+Zn^(2+) (aq), E^(Theta) = 1.20V
The equilibrium constant of the reaction : Zn(s)+2Ag^(+)(aq)toZn(aq)+2Ag(s),E^(@)=1.50V at 298 K is
Calculate the equilibrium costant log (K_(c)) for the reaction Zn(s)+Cu^(2+)(aq)rarrZn^(2+)(aq)+Cu(s) [Given E_(cell)^(2)=1.1 V ]
