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
An aqueous solution containing one mole per litre of each Cu(NO_3)_2 , AgNO_3 . Hg(NO_3)_(2), Mg(NO_3)_2 is being electrolysed using inert electrodes. The values of standard electrode potential (reduction potential) in volts are Ag//Ag^(+) = + 0.80 V " "Hg//Hg^(2+) = + 0.79V " " Cu//Cu^(2+) = +0.34 V " " Mg//Mg^(2+) = -2.37 V With increasing voltage, the sequence of deposition of metals on cathode will be
If K_(c) for the reaction Cu_((aq))^(2+) + Sn_((aq))^(2+) to Sn_((aq))^(4+) + Cu_((s)) at 25^(@) C is represented as y xx 10^(6) then find the value of y . (Given : E(Cu^(2+) |Cu)^(@) = 0.34 V , E_(Sn^(4+)|Sn^(2+))^(@) = 0.15V )
Calculate E of the cell with Cu^(2+) (0.1)//Cu electrode and Cl^(-)(0.1M)// 1/2Cl_(2) , Pt electrode. E^(@) Cu^(2+)//Cu= +0.33V and E^(@) 1/2 Cl_(2)//Cl^(-) = 1.36V
Calculate the Electrode potential of single electrode. Cu^(2+)(0.01M)//Cu" "(E^(@)= +0.337V)
The equilibrium constant of the reaction Cu_((s)) + 2Ag_((aq))^(+) iff Cu_((aq))^(+2) + 2Ag_((s)) , " is " x xx 10^(15) , 'x' is ( E^(@) = 0.46 V at 298 K )
Cu^+ ion is not stable in aqueous solution because of diproportionation reaction . E^(@) value for the disproportionation of Cu^(+) is : (E_(cu^(2+)// cu^+)^(@) = + 0.15 V , E_(cu^(+) // cu)^(@)= + 0.34 V )
