[" Consider the following four electrodes "],[A=Cu^(2)(0000m)/Cu_(10)B=Cu^(2)(0.1M)/Cu_(10)],[C=Cu^(2)(0.0mm)Cu,D=Cu^(2)(0.00m)/Cu_(2)],[" If the standard reduction potential of "Cu^(-2)/Cu" is "+0.34V" ,the reduction "],[" potential (in volts) of the above electrodes follow the order "]

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

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

Consider the following four electrodes : P=Cu^(2+)"(0.001 M)|Cu(s) Q=Cu^(2+)("(0.1 M)"|Cu(s) R=Cu^(2+)"(0.01 M)"|Cu(s) S=Cu^(2+)"(0.001)"|Cu(s) If the standard electrode potential of Cu^(2+)|Cu is +0.34V , the reduction potential in volts of the above electrodes follows the order-

If standard electrode potenial of Cu^(2+)//Cu is 0.34V then potential of Cu dipped in 0.1 M solution of CuSO_(4) will be

The standard reduction potential for Cu^(2+)|Cu is +0.34V. Calculate the reduction potential at pH=14 for above couple. K_(sp) of Cu(OH)_2 is 1.0xx10^-19 .

The standard reduction potential for Cu^(2+)//Cu is +0.34V. If ksp of Cu(OH)_2 is 1.0 xx 10^(-19) , the reduction potential at pH =14 for the above couple is