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Given that the standard electrode (E^(@)...

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.

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To arrange the given metals in increasing order of their reducing power based on their standard electrode potentials, we can follow these steps: ### Step 1: Understand the Relationship between Standard Electrode Potential and Reducing Power The standard electrode potential (E°) indicates how easily a species can be reduced. A more negative E° value means that the species is more easily oxidized, thus acting as a stronger reducing agent. Therefore, the reducing power increases as the standard electrode potential becomes more negative. ### Step 2: List the Given Standard Electrode Potentials We have the following standard electrode potentials: - \( K^{+} // K = -2.93 \, V \) ...
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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 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.

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

Given E_(Ag^(+)//Ag)^(@)=0.80V , E_(Mg^(2+)//Mg)^(@)=-2.37V , E_(Cu^(2+)//Cu)^(@)=0.79 E_(Hg^(2+)//Hg)^(@)=1.71V Which of the following statements is/are correct?

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