Four metals W, X, Y and Z have the following values of `E_("red")^@` :
`E_("red")^(@)`
W = - 0.140 V
X = - 2.93 V
Y = +0.80 V
Z = +1.50 V
Arrange them in the increasing order of reducing power.

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

The standard E_(red)^(@) Values if A, B and C are + 0.68V , - 0.76V , -0.50V respectively. The order of their reducing power is :

Reduction potentials of four elements P, Q, R, S is -2.90V, 0.34V, 1.2V and -0.76V . The decreasing order of reducing power is

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

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

Given that the standard elctrode potentials (E^(Theta)) of metals are: K^(+)//K= -2.93V, Ag^(+)//g=0.80V , Cu^(2+)//Cu=0.34V, Mg^(2+)//Mg=-2.37V , Cr^(3+)//Cr= -0.74V, Fe^(2+)//Fe= -0.44V . Arrange these metals in an increasing order of their reducing power.

Given that E^o values of Ag^+//Ag, K^+//K, Mg^(2+)//Mg and Cr^(3+) //Cr are 0.08 V, -2 .93 V, -2 37 V and -0. 74 V respectively. Therefore the order for the reducing power of the metal is .

Which of the following is correct if E_(x^+/x)^0 >E_(y^+/y)^0>E_(z^+/z)^0

The half cell reactions with reduction potentials are Pb(s) to Pb^(+2) (aq), E_("red")^(@) =-0.13 V Ag(s) to Ag^(+) (aq) , E_("rod")^(@) =+0.80 V Calculate its emf.

In the following circuit , E_1 = 4V, R_1 = 2 Omega, E_2 = 6 V, R_2 = 2Omega, and R_3 = 4Omega. Find the currents i_1 and i_2