What is `E^(c)._(red)` for the reaction`: Cu^(2+)+2e^(-) rarr Cu` in the half cell `Pt_(S^(2)|CuS|Cu)`. if `E^(c-)._(Cu^(2+)|Cu)` is `0.34V` and `K_(sp)` of `CuS=10^(-35)`?

Calculate the half-cell potential at 298 K for the reaction Cu^(2+) ( aq) + 2e^(-) rarr Cu(s) where [ Cu^(2+) ] is 5.0 M and E^(@) is + 0.34 V

Consider the cell reaction : Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq) If E^(c-)._(Mg^(2+)|Mg(s)) and E^(c-)._(Cu^(2+)|Cu(s)) are -2.37 and 0.34V , respectively. E^(c-)._(cell) is

Consider the cell reaction : Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq) If E^(c-)._(Mg^(2+)|Mg(s)) and E^(c-)._(Cu^(2+)|Cu(s)) are -2.37 and 0.34V , respectively. E^(c-)._(cell) is

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

For the disproportion of copper: 2 Cu^(+) to Cu^(+2) + Cu E^(0) is :- Given E^(0) for Cu^(+2)//Cu is 0.34 V & E^(0) for Cu^(+2)//Cu^(+) is 0.15 V:

For the disproportion of copper: 2 Cu^(+) to Cu^(+2) + Cu E^(0) is :- Given E^(0) for Cu^(+2)//Cu is 0.34 V & E^(0) for Cu^(+2)//Cu^(+) is 0.15 V: