For the galvanic cell, `Cu|Cu^(2+)(0.13M)||Ag^(+)(0.01M)|Ag`
(i) Determine the reduction potential of each electrode and the EMF of the cell.

Consider the galvanic cell Cu|Cu^(2+)(0.13M)||Ag^(+)(0.01M)|Ag (i) Calculate the reduction potential of each electrode and EMF of the cell (ii) Is is reaction representing the cell spontaneous ? ["Given : "E_(Cu^(2+)|Cu)^(@)=+0.34V and E_(Ag^(+)|Ag)^(@)=+0.80V]

For the galvanic cell, Cu|Cu^(2+)(0.13M)||Ag^(+)(0.01M)|Ag (ii) Is the cell reaction spontaneous ? ["Given : "E_(Cu^(2+)|Cu)^(@)=+0.34V and E_(Ag^(+)|Ag)^(@)=+0.80V] .

For the galvanic cell : Ag|AgCl(s), KCl (0.2M)||KBr(0.001M), AgBr(s)|Ag, calculate the EMF and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at 25^(@)C . Given : K_(sp)(AgCl)=2.8xx10^(-10), K_(sp)(AgBr)=3.3xx10^(-13)

For the cell, Cu|Cu^(2+)||Hg^(2+)|Hg, E_("cell")^(@)=+0.45V . Find the standard reduction potential of Hg^(2+)|Hg electrode. Given : E_(Cu^(2+)|Cu)^(@)=+0.34V.

At 25^(@)C, E_("cell") of the galvanic cell Zn|Zn^(2+)(aq, 0.01M)||Cu^(2+)(aq, 0.01M)|Cu is +1.07V . The equilibrium constant of the cell reaction occurring in this cell is -

For the cell Cu//Cu^(2+)//Hg^(2+)//Hg , E_(cell)^(6) = 0.45 V . Find the standard reduction potential of Hg^(2+)//Hg electrode. Given E^@ cu^(2+)//cu = 0.34 .

Consider the following galvanic cell given below- Cu|Cu^(2+)||Cl^(-)|Cl_(2) , Pt. Write the reactions that occur at anode and cathode.

For a standard cell, Cu(s)|Cu^(2+)(0.01M)||Ag^(+)(0.02M)|Ag(s) E_(Cu^(2+)|Cu)^0=+0.34V,E_(Ag^+//Ag)^0=+0.80V Write the reactions taking place at the electrodes.

For the half - cell Fe^(3+), Fe^(2+)|Pt , the concentration of Fe^(2+) and Fe^(3+) ions are 0.1 and 0.01 (M) respectively. Determine the reduction potential of the half - cell at 25^(@)C . Given : E_(Fe^(3+), Fe^(2+)|Pt)^(@)=+0.77V