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Calculate Delta(r)G^(@) and e.m.f. (E) t...

Calculate `Delta_(r)G^(@)` and e.m.f. (E) that can be obtained from the following cell under the standard conditions at `25^(@)C: Zn(s)|Zn^(2+)(aq)" "Sn^(2+)(aq)|Sn(s)`
Given `E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Sn^(2+)//Sn)^(@)=-0.14V`
And `F=96500C" "mol^(-1)`
OR
(a) Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
(b) Calculation the standard cell potential of the galvanic cell in which the following reaction takes place: `Fe^(2+)(aq)+Ag^(+)(aq)toFe^(3+)(aq)+Ag(s)`
Calculate the `Delta_(r)G^(@)` and equilibrium constant of the reaction also.
`(E_(Ag^(+)//Ag)^(@)=0.80,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V)`

Text Solution

Verified by Experts

In the given equation n=2,
`F=96500" C "mol^(-1)`
`E_(cell)^(@)=-0.14V`
Therefore, `DeltarG^(@)=-2xx-0.14Vxx96500xx" C "mol^(-1)=0.28xx96500=27020Kj" "mol^(1)`
`E_(cell)^(@)=E^(@)" cathode"-E_("anode")^(@)`
`E_(cell)^(@)=-0.14V-(-0.76)=0.14V+0.76=0.62V`
`DeltaG^(@)=n" F E"_(cell)^(@)`
`=-2xx96500xx0.62=-1.24xx96500=-119600=-1.19660xx10^(5)J//mol`
OR
(a) Conductivity: Conductivity of a solution is equal to the conductance of a solution of 1 cm length and cross section area of 1 cm length and cross section area of 1 square cm. it may also be define as the conductance of in centimeter cube of the conductor. it is represented by the symbol kappa(K)
`K=1//P`
e is resistivity. the unit of K is `ohm^(-1)cm^(-1)` or S `cm^(-1)`. The conductivity,K, of a electrolytic solution depends on the concentration of the electrolyte. Natre of solvent and temperature.
Molar conductivity: Molar conductivity of a solution at given concentration is the conductance of the volume V of solution containing one mole of electrlyte Kept between two electrodes with area of cross section A and distance of Unit length. therefore, distance is Unit so
1=1
Volume=Area of bare`xx`length
So, `V=Axx1`
`=A`
`^^_(m)=KA//1`
`^^_(m)=KV`
(b) `E_(cell)^(@)=+0.80V-0.77V=+0.03V`
`DeltarG^(@)=nFE_(cell)^(@)`
`=-1(mol)xx(96500" C "mol^(-1))xx(0.03V)`
`=-2895" CV "mol^(-1)=-2895" S "mol^(-1)`
`DeltarG^(2)=-2.303RT" log "K_(C)`
`=-2895=-2.303xx8.314xx298xxlog" "K_(C)`
`logK_(C)=0.5074`
`K_(C)="Antilog"(0-5074)`
`K_(C)=3.22`
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