`E^(@)(Ni^(2+)//Ni) =- 0.25` volt, `E^(@)(Au^(3+)//Au) = 1.50` volt. The emf of the voltaic cell `Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au` is:-

E^(@) (Ni^(2+)//NI) = -0.25 "volt", E^(@) (Au^(3+)//Au) = 1.50 "volt" . The emf of the voltaic cell, Ni|Ni^(2+) (1.0M) || Au^(3+) (1.0 M)|Au is :

The emf of the cell, Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@) for Ni^(2+)//Ni =- 0.25 volt, E^(@) for Ag^(+)//Ag = 0.80 volt] is given by : [E^(@) for Ag^(+)//Ag = 0.80 volt]

The emf of the cell, Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@) for Ni^(2+)//Ni =- 0.25 volt, E^(@) for Ag^(+)//Ag = 0.80 volt] is given by : [E^(@) for Ag^(+)//Ag = 0.80 volt]

The emf of the cell, Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@) for Ni^(2+)//Ni =- 0.25 volt, E^(@) for Ag^(+)//Ag = 0.80 volt] is given by : [E^(@) for Ag^(+)//Ag = 0.80 volt]

Ni|Ni^(2+)(1.0M)||Au^(3+)(1.0M)| Au (where E^(@) for Ni^(2+)//Niis -0.25and V and E^(@) for Au^(3+)//Au is (0.150V). What is the emf of the cell ?

These question consist of two statements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Ni//Ni^(2+) (1.0 M) || Au^(3+) (1.0 M) | Au , for this cell emf is 1. 75 V if E_(Au^(3+)//Au)^@ =1.50 and E_(Ni^(3+)//Ni)^2 =0.25 V . Emf of the cell =E_("cathode")^@- E_("anode")^@ .

E^(@) values of Ni^(2+) , Ni and Cl_(2), Cl^(-) are respectively -0.25V and +1.37V . Calculate the EMF of the cell Ni, Ni^(2+) (0.01M)//Cl^(-)(0.1M), Cl_(2) , pt. Potential of nickel electrode is given as,