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Arrange Ag, Cr and Hg metals in the incr...

Arrange Ag, Cr and Hg metals in the increasing order of reducing power. Given :
`E_(Ag^(+)//Ag)^@ = + 0*80 V`
`E_(cr^(+3)//Cr)^@ = -0*74 V`
`E_(Hg^(+2)//Hg)^@ = +0*79 V`

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To arrange the metals Ag (Silver), Cr (Chromium), and Hg (Mercury) in the increasing order of reducing power, we will use their standard reduction potentials (E° values). The standard reduction potentials provided are: - \( E^\circ_{Ag^+ / Ag} = +0.80 \, V \) - \( E^\circ_{Cr^{3+} / Cr} = -0.74 \, V \) - \( E^\circ_{Hg^{2+} / Hg} = +0.79 \, V \) ### Step-by-Step Solution: ...
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Arrange the following metals in increasing order of their reducing power. E_(K^(+)//K)^(@)=-2.93V, E_(Ag^(+)//Ag)^(@)=+0.80V, E_(Al^(3)//Al)^(@)=-1.66V E_(Au^(3+)//Au)^(@)=+1.40 V, E_(Li^(+)//Li)^(@)=-3.05 V

Four different solution containing 1M each of Au^(+3), Cu^(+2), Ag^(+), Li^(+) are being electrolysed by using inert electrodes. In how many samples, metal ions would be deposited at cathode? ["Given :" E_(Ag^+//Ag)^(0) = 0.8, E_(Au^(+3)//Au)^(0) = 1.00 V E_(Cu^(+2)//Cu)^(0) = 0.34 V, E_(Li^(+)//Li)^(0)= -3.03 V ]

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:

(i) On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti^(4+) species may be used to oxidise Fe(II) to Fe(III) {:(Ti^(4+) + e^(-) to Ti^(3+), E^(@) = +0.01V), (Fe^(3+) + e^(-) to Fe^(2+), E^(@)= +0.77V):} (ii) Based on the data arrange Fe^(3+), Mn^(2+) " and " Cr^(2+) in the increasing order of stability of +2 oxidation state. (Give a brief reason) E_(Cr^(3+)//Cr^(2+))^(@) = -0.4V E_(Mn^(3+)//Mn^(2+))^(@) = +1.5V E_(Fe^(3+)//Fe^(2+))^(@) = +0.8V

An aqueous solution containing 1M each of Au^(3+),Cu^(2+),Ag^(+),Li^(+) is being electrolysed by using inert electrodes. The value of standard potentials are : E_(Ag^(+)//Ag)^(@)=0.80 V,E_(Cu^(+)//Cu)^(@)=0.34V and E_(Au^(+3)//Au)^(@)=1.50,E_(Li^(+)//Li)^(@)=-3.03V will increasing voltage, the sequence of deposition of metals on the cathode will be :

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.

Under standered condition Delta G^(@) for the reaction 2Cr(s)= 3Cd^(2+)(aq) rarr 2Cr_((a a))^(3+) +3Cd(s) is (E_(Cr^(3+)//Cr)^(@) = - 0.74 V, E_(Cd^(2+)//Cd)^(@) = - 0.4 V)

Determine range of E^(@) values for this reaction X_(aq)^(2+)+2e^(-)toX(s) for given conditions: (a). If the metal X dissolve in HNO_(3) but not in HCl it can displace Ag^(+) ion but not Cu^(2+) ion. (b). If te metal X is HCl acid producing H_(2)(g) but does not displace either Zn^(2+) or Fe^(2+) Given E_(Ag^(+)//Ag)^(0)=0.8V" "E_(Fe^(2+)//Fe)^(0)=-0.44 E_(Cu^(2+)//Cu)^(0)=0.34V" "E_(NO_(3)^(-)//NO)^(0)=0.96V" "E_(zn^(2+)//Zn)^(0)=-0.76V

Calculate standard free energy change for the reaction 2Ag+2H^(+)rarrH_(2)+2Ag^(+) Given : E_(Ag^(+)//Ag)^(@)=+0.80V

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