The voltage of a certain cell has standard potential at `25^(@)C` and `20^(@)C` are 0.3525 V and 0.3533 V respectively. If the number of electrons involved in te overall reactions are two, calculate `triangleG^(@), triangleS^(@)` and `triangleH^(@)` at `25^(@)C`.

Answer

`triangleS^(0)=-30.88JK^(-1)`
`triangleH^(0)=-77.23kJ`
`triangleG^(0)=-68.03kJ`

Explanation

`E_(25^(@)C)^(0)=0.3525"volt" " "E_(20^(@)C)^(0)=0.3533"volt" " "n=2`
`(dE)/(dT)=((E_(25^(@)C)^(0)-E_(20^(@)C)^(0))/(T_(2)-T_(1)))=(0.3525-0.3533)/(25-20)=-1.6xx10^(-4)"volt"deg^(-1)`
`triangleG_(25)^(@)=triangleH_(25)^(@)-298xxtriangleS_(25)^(@)=-2xx0.3525xx96500=-68.03kJ`
`triangleS_(@)=2xx96500=-68.03kJ`
`triangleS^(@)=2xx96500xx(-1.6xx10^(-4))=-30.88Jk`
`triangleH^(@)=triangleG^(@)+TtriangleS^(@)=-68.03-298xx(30.88)/(1000)=-77.23kJ`