The standard potential of the following cell is 0.23 V at `15^@C and 0.21V` at `35^@C`
`(Pt) H_2(g) |HCl(aq)|AgCl(s)|Ag(s)`
Calculate `Delta H^@ and Delta S^@` for the cell reaction by assuming that these quantities remain unchanged in the range `15^@` to `35^@C`

Explanation

`(DeltaG^@)_(298)=-nFE^@=-1 times 96500times 0.23=-22195 J =-22.195 kJ`
`(Delta G^@)_(398)=-1 times 96500 times 0.21=-20265 J=-20.265 kJ`
Now we have
`therefore -22.195 =Delta H^@-288 Delta S^@`
and `-20.265 =Delta H^@-308.Delta S^@`
for these equations, we get
`Delta H^@=49.98 kJ`
`Delta S^@=-96.50kJ`