Calculate the emf of the cell at 298K `E_(NI^(2+)//NI)^0 is -0.25V and E_(Ag^+//Ag)^0 is +0.80V`.

Calculate the e.m.f of the following concentration cell at 298K, Zn(s)| Zn^(2+)(0.1) || Zn^(2+) (1.0)|Zn(s)

A substance that will reduce Ag to Ag but will not reduce Ni^(2+) to Ni is : (E_(Ag^(+)//Ag)^(0))=0.80 V, E_(Ni^(2+)//Ni)^(0)

Calculate the emf of the following cell at 25^@C Fe|FeSO_4||CuSO_4|Cu (0.1 M) (0.01 M) Given that E^@ (oxd.) of Fe and Cu are 0.44V and -0.34V respectively.

Calculate the emf of the cell in which the following reaction takes place: Ni(s)+ 2Ag^+(0.002M) to Ni^2(0.160 M)+2Ag(s) . Given that E_(cell)^0=1.05V

At equimolar concentration of Fe^(2+) and Fe^(3+) what must [Ag^(+)] be so that the voltage of the galvanic cell made from Ag^(+)//Ag and Fe^(3+)//Fe^(2+) electrodes equals zero? The reaction is Fe^(2+) + Ag^(+) hArr Fe^(3+) + Ag . Determine the equilibrium constant at 25^(@)C for the reaction. Given E_(Ag^(+)//Ag)^(@)= 0.799V and E_(Fe//Fe^(2+))^(@)= 0.771V

Write the Nernst equation and emf of the following cells at 298 K. Sn_(s)//Sn^(2+) (0.050M) II H^+ (0.02M)//H_2(g) (1 bar) 1 Pt(S), E^0 Sn^(2+)//Sn=-0.14 V .

Calculate the reduction potentials for the following half cells: Ag|Ag^+ (10^-5 M),E_(Ag^+,Ag)^@=0.80V

The e.m.f a cell is 0.265V at 25^(@)C and 0.2594V at 35^(@)C . Hence

An aqueous solution containing 1M each of Au^(3+), Cu^(2+), Ag^(+), Li^(+) is being electrolysed by using inert electrodes. The value of standard potentials are E_(Ag^(+)//Ag)^(@)= 0.80V, E_(Cu^(2+)//Cu)^(@) = 0.34V, and E_(Au^(3+)//Au)^(@) = 1.50 V , E_(Li^(+)//Li)^(@)= -3.03V , With increasing voltage, the sequence of deposition of metals on the cathode will be