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Given the standard electrode potentials ...

Given the standard electrode potentials ,
`K^(+)//K = - 2.93 V , Ag^(+) // 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 the reducing power .

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The correct Answer is:
`Ag
Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, the increasing order of reducing power will be `AgltHgltCrltMgltK`.
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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.

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.

The potential associated. with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gås appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further the reaction is carried out at 298 K, then the potential of each electrode is said to be the standard electrode potential. By convention, the standard electrode potential of hydrogen electrode is 0:0 volt. The electrode potential value for each electrode process is a measure, of relative tendency of the active species in the process to remain in the oxidized / reduced form. A negative E^@ means that the redox couple is a stronger reducing agent than the H^(+)//H_2 couple. A positive E mears that the redox couple is a weaker reducing agent than. the H^(+)//H couple. The metal with greater positive value of standard reduction potentlal forms the oxide of greater thermal stability: Given the standard reduction potentials. E_(K^(+)//K)^(@)=-2.93V, E_(Ag^(+)//Ag)^(@)=+0.80V, E_(Hg^(+)//Hg)^(@)=0.79V E_(Mg^(+)//Mg)^(@)=-2.37V, E_(Cr^(3+)//Cr)^(@)=-0.74V The correct increasing order of reducing power is:

Given the standard electrods potential K^(+)//K=-3.02V Cu^(+2)//Hg=+0.34V Hg^(+2)//Hg=0.92 Cr^(+3)//Cr=-0.74V Decreasing order of reducing power of these element is

The values of some of the standard electrode potential are E^(@)(Ag^(+)//Ag)=0.80 V,E^(@)(Hg_(21+)^(2+)//Hg)=0.79 VE^(@)(Cu^(+2)//Cu)=0.34V What is the sequence of deposition of metals on the cathode ?

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.

The standard electrode potential a Ag^(+)//Ag is +0.80 V and of Cu^(2+)//Cu is +0.34 V. These electrodes are connected through a salt bridge and if :

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