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Given standard electrode potentials K^...

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.

Text Solution

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Lower the electrode potential better is the reducing agent since the electrode potential increase in the order `K^(+)//K(-2.93 V)Mg^(2+)//Mg(-2.37 V),Cr^(3+)//Cr(-074 V),F^(2+)//Hg(0.79V ),Ag^(+)//Ag(0.80 V)` therefore oreducing power of metals decrases in the same order i.e k, Mg , Cr Hg ,Ag
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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 is their increasing order of reducing power .

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

Knowledge Check

  • Given 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 Which acts as a better oxidising agent?

    A
    `Mg^(2+)`
    B
    `Cr^(3+)`
    C
    `Ag^(+)`
    D
    `k^+`

  • 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:

    A
    `K gt Mg gt Cr gt Hg gt Ag`
    B
    `Ag gt Hg gt Cr gt Mg gt K`
    C
    `Mg gt K gt Cr gt Hg gt Ag`
    D
    `Cr gt Hg gt K gt Mg gt Ag`

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