At 298 K the standard free energy of formation of `H_(2)O(l)` is `-237.20 kJ/"mole"` while that of its ionisation into `H^(+)` ions and hydroxyl ions is `80kJ/"mole"`, then the emf of the following cell at 298 K will be :
[Take 1F = 96500 C]
`H_(2)O(g, 1 bar)|H^(+)(1M)||OH^(-)(1M) |O_2)(g, 1bar)`

Options

A. `0.40` V
B. `0.81` V
C. `1.23 V `
D. `0.40` V

Answer

A

Explanation

Cell reaction
cathode :

Also we have
`H_(2) (g) + (1)/(2) O_(2) (g) to H_(2) O (l),`
`Delta G_(f)^(@) = -237.2 kJ`/mole
`H_(2)O (l) to H^(+) (aq) + OH^(-) (aq)`
`Delta G^(@) = 80` kJ/mol
Hence for cell reaction
`Delta G^(@) = -237.2 + (2 xx 80) = -77.20` kJ/mol
`therefore E^(@) = - (DeltaG^(@))/(nF) = (77200)/(2 xx 96500) xx 0.40` V