Write the cell reactions for
a) `Ni//Ni^(2+)// //Cu^(2+)//Cu`
b) `Cu^(2+)//Cu"//"Cl^(-) 1/2 Cl_(2)`

Write the configuration of Cut, Cu^(+), Co^(2+), Mn^(2+),

The emf of the following three galvanic cells are respresented by E_(1), E_(2) and E_(3) respectively. Which of the following is correct? (i) Zn//Zn^(2+) (1M)// //Cu^(2+)(1M)//Cu (ii) Zn//Zn^(2+) (0.1M)// //Cu^(2+)(1M)//Cu (iii) Zn//Zn^(2+)(1M)// //Cu^(2+)(0.1M)//Cu

The cell reaction of the galvanic cell, Cu(s) // Cu^(2+)(aq) // Hg^(2+) (aq) // Hg (l) is

The e.m.f of the cell, Ni//Ni^(2+)(1M)////Cl^(-)(1M)//Cl_(2),Pt is E^(@)Ni^(2+)//Ni= -0.25V,E^(@)1//2Cl_(2)//Cl^(-)=1.36V

Consider the following four electrode : A = Cu^(2+) (0.0001 M)// Cu(S) B= Cu^(2+) (0.1M)//Cu(s) C = Cu^(2+) (0.01 M )//Cu(s) D) = Cu^(2+) (0.001 M )// Cu(s) If the standard reduction potential of Cu^(+2)// Cu is + 0.34 V , the reduction potential ( in volts ) of the above electrodes follow the order

Calculate the EMF of the cell Cu//Cu^(2+)(0.1M)"//"Cl^(-) (0.01)// 1/2 Cl_(2) , Pt E^(@) Cu^(2+)//Cu = +0.33V E^(@) 1/2 Cl_(2)//Cl^(-) = +1.36V

E.M.F of the cell reaction , 2Ag^(+) + Cu to 2Ag + Cu^(2+) is 0.46 V . If E_(Cu^(2+) // Cu)^(0) is | 0.34 V , E_(Ag^(+)//Ag)^(0) is

If the Pb^(2+) concentration is maintained at 1.0M what is the [Cu^(2+)] , when the cell potential drops to zero ? E_("cell")^@ = 0.473V, Pb(s) // Pb^(2+) // Cu^(2+) // Cu(s)