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

Given the standard electrode potential value of some metals K^+//K = -2.93V,Ag^+//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 the standard electrode potentials: k^+//k = -2.93v , Ag^+//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 the standard electroge potentials. K^+//K= -2.93V,Ag^+//Ag=0.80V, Mg^(2+)//Mg=-2.37V . Arranges these metals in their decreasing order of oxidising power.

Following are the data of standard reduction half cell potentials of certain electrochemical cells. E_(Ag^(+)//Ag)^@ = 0.80V & E_(Cu^(2+)//Cu)^@ = 0.34V E_(Sn^(2+)//Sn)^@ =-0.14V & E_(Na^(2+)//Na)^@ = -2.74V Also identify andode, cathode and direction of flow of current in each of the constructed cells.

Following are the data of standard reduction half cell potentials of certain electrochemical cells. E_(Ag^(+)//Ag)^@ = 0.80V & E_(Cu^(2+)//Cu)^@ = 0.34V E_(Sn^(2+)//Sn)^@ =-0.14V & E_(Na^(2+)//Na)^@ = -2.71V Represent the electrochemical cell constructed from these pair of half cells.

Following are the data of standard reduction half cell potentials of certain electrochemical cells. E_(Ag^(+)//Ag)^@ = 0.80V & E_(Cu^(2+)//Cu)^@ = 0.34V E_(Sn^(2+)//Sn)^@ =-0.14V & E_(Na^(2+)//Na)^@ = -2.73V Calculate the standard EMF of the cells.

The standard reduction potential values of three metal cations, X, Y and Z are +0.52, -3.03, -1/18V respectively. Arrange the corresponding metals in order of their increasing reducing power.

Which is a stronger reducing agent Zn or Fe? If E_(Zn^(2+)//Zn)^@ = -0.76V , and E_(Fe^(2+)//Fe = -0.44V

Fe^(3+)+e rarr Fe^(2+)_(aq) , E^@ = -0.77V Write the reaction which could be feasible using above half cells.

The electrode potentials for the two half cell reaction is given below. Determine emf Mg^(2+)+2e^- rarr Mg(s), E=-2.37V Cu^(2+)+2e^- rarr Cu(s), E=+0.34V.