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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.

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To determine the increasing order of reducing power for the given metals based on their standard electrode potentials, we will follow these steps: ### Step 1: Understand the Concept of Reducing Power Reducing power is the ability of a substance to donate electrons and reduce another substance. A substance with a lower (more negative) standard electrode potential is a stronger reducing agent because it is more likely to lose electrons (get oxidized). ### 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 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 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:

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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 ?

A solution containing 1 mol per litre of each Cu(NO_(3))_(2), AgNO_(3), and Hg_(2)(NO_(3))_(2) is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts ( reduction potential ) are Ag|Ag^(o+)=+0.80, 2Hg|Hg_(2)^(2+)=+0.79 Cu|Cu^(2+)=+0.34,Mg|Mg^(2+)=-2.37 . With increasing voltage, the sequence of deposition of metals at the cathode will be

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.

A solution containing one mole per litre each of Cu(NO_(3))_(2),AgNO_(3),Hg(NO_(3))_(2) and Mg(NO_(3))_(2) is being electrolysed by using inerty electrodes. The values of the standard oxidation potentials in vlts are Ag//Ag^(+)=-0.8V,Ag//Hg^(2+)=-79V,Cu//Cu^(2+)=-0.34V,Mg//Mg^(2+)=2.37V . The order in which metals will be formed at cathode will be-

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Given : E^(c-)._(Ag^(o+)|Ag)=0.80V, E^(c-)._(Mg^(2+)|Mg)=-2.37V, E^(c-)._(Cu^(2+)|Cu)=0.34V,E^(c-)._(Hg^(2+)|Hg)=0.79V Which of the following statements is // are incorrect ?

NCERT ENGLISH-REDOX REACTIONS-EXERCISE
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