Construct the cell corresponding to the reaction : `3Cr^(2+)(1M)rarr 2Cr^(3+)(1M)+Cr(s)` and predict if the reaction is spontaneous. Also calculate the following
(i) `Delta H` and `Delta S` of the reaction at `25^(@)C`. Given `E_(Cr^(3+),Cr)^(@)=0.5 V, E_(Cr^(3+),Cr^(2+))=-0.41 V`
`Delta G` of the reaction at `35^(@)C =-270.50 kJ`

Options

A. `Delta H=-51.05 kJ, Delta S =700 k JK^(-1)`
B. `Delta H=-31.05 kJ, Delta S = 700 kJ K^(-1)`
C. `Delta H =-51.05 kJ, Delta S=706 kJK^(-1)`
D. `Delta H = - 53.05 kJ, Delta S = 706 k J K^(-1)`

Answer

D

Explanation

The cell corresponding to given reaction is as follows.
`Cr^(2+)(1M)|Cr^(3+)(1M)||Cr^(2+)(1M)|Cr(s)`
`I Cr^(3+)+3e rarr Cr(s), Delta G_(1)^(@)=-3FE_(Cr^(3+),Cr)^(@)`
`II Cr^(3+)+e rarr Cr^(2+), Delta G_(II)^(@)=-FE_(Cr^(3+),Cr^(2+))^(@)`
Eq. I- Eq. II : `Cr^(2+)+2e rarr Cr(s), Delta G^(@)=-3xx0.5F-0.41 F=-1.91 F`
`therefore E_(Cr^(2+),Cr)^(@)=(1.91)/(2)=0.955 V`
`E_("cell")^(@)=E_(Cr^(2+),Cr)^(@)-E_(Cr^(3+),Cr^(2+))^(@)=0.955+0.41=1.365V`
`E_("cell")^(@)` is `+ve` so `Delta G^(@)` is -Ve and hence the given reaction is spontaneous
(i) `Delta G^(@)=-nF E_("cell")^(@)=-2xx96500xx1.365 J=-263.44 kJ`
From Gibb s - Helomholtz equation
`Dela G=Delta H+T[(del(Delta G))/(del T)]_(P)-263.44=Delta H+298((-270.50+263.44))/(10)`
`=Delta H-298xx0.706 therefore Delta H=-53.05 kJ`
`Delta S=(Delta H-Delta G)/(T)=(-53.05+263.44)/(298)=0.706 kJ`
`K^(-1)=706 kJ K^(-1)`.