Find the cell potential at `25^@C` for the cell-`Mg//Mg^(2+)(0.001)//Cu^(2+)(0.0001 M)//Cu` Given `E^@(Mg^(2+)//Mg) = - 2.37V`, `E^@(Cu^(2+)//Cu) =+0.34V`

Calculate the emf at 25^@C for the cell Mg(s)|Mg^(2+) (0.01M)||Sn^(2+)(0.1M)|Sn(s) (Given E_(Mg^(2+)|Mg)^0 = -2.34V & E_(Sn^(2+)|Sn)^0 = -0.136V)

Write the cell reaction and cell potential of the cell Zn//Zn^(2+)(0.01 M)abs"Ag^+(0.1M)//Ag. Given E^0(Zn^(2+)//Zn)=-0.76V E^0(Ag^+//Ag)=+0.80V .

The standard cell potential for the cell is , Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu [E^@ for Zn^(2+)//Zn=-0.76V, E^@ for Cu^(2+)//Cu=+0.34V]

Discuss Daniel cell with cell reaction. Write the cell reaction.Write the cell reaction and find the emf of the cell. Mg | Mg^2+ || Cu^2+ | Cu at 25^@ Given E^@ Mg^2+ | Mg = -2.37 V (0.001 M) 0.0001M and E^@ Cu^2+ // Cu = 0.34V .

Calculate the half cell potential at 25^@C for the cell reaction Cu^(2+)(aq)+2erarrCu , When [Cu^(2+)]=4M and E^@(Cu^(2+)//Cu)=0.34V .

A cell is set up between ‘Zn’ and ‘Cu’ electrode. If the two half cells work under standard condition, calculate the cell potential. Given E^@(Zn^(2+)//Zn)=-0.76V and E^@(Cu^(2+)//Cu)=+0.34V .

Given the standard electrode potentials, K^(+) // K = - 2.93 V, Ag^(2+) //Ag = 0.80 V Hg^(2+) //Hg = 0.79 V , Mg^(2+) //Mg = - 2.37 V, Cr^(3+) //Cr = - 0.74 V Arrange these metals in their increasing order of reducing power.

Write the Nernst equation and emf of the following cell at 298 K. (i) Mg(s) |Mg^(2+) ( 0.001 M) || Cu^(2+) ( 0.0001M ) | Cu(s) (ii) Fe(s) | Fe^(2+) ( 0.001M) || H^(+) ( 1M) | H_(2)(g) (1 "bar") | Pt(s) Given that, E_(Mg^(2+) //Mg)^(@) = - 2.36V , E_(Cu^(2+) Cu)^(@) = 0.34V, E_(Fe^(2+)Fe)^(@) = -0.44V

Calculate the standard electrode potential of Ni^(2+) | Ni electrode if emf of the cell, Ni(s) | Ni^(2+) ( 0.01 M ) || Cu^(2+) ( 0.1M) | Cu(s) is 0.059 V. [ Given E_(Cu^(2+) //Cu) ^(@) = + 0.34V