Fuel Cells: Fuel cells aare galvanic cells in which chemical energy of fuel is directly converted into electrical energy.

A type of fuel cells is a hydrogen-oxygen fuel cell. It is a consists of two elecrtodes made up of two porous graphite impregnated with catayst (platinum, silver or metal oxide). the electrode are place in aqueous solution of `NaOH`. Oxygen and hydrogen are continuously fed into the cell as sow in Fig. Hydrogen gets oxidised to `H^(+)` which is nautralised by `OH^(-)` i.e., Anodic reaction.
`H_(2) hArr 2H^(+) +2e^(-)`
`2H^(+) +2OH^(-) hArr 2H_(2)O`
`H_(2) +2OH^(-) hArr 2H_(2)O +2e^(-)`
At cathode `O_(2)` gets reduced to `OH^(-)`
i.e., `O_(2) +2H_(2)O +4e^(-) hArr 4OH^(-)`
Hence the net reaction is
`2H_(2)+O_(2) hArr 2H_(2)O`
The overall reaction has `DeltaH =- 285.6 kJ//mol` and `DeltaG =- 237.4 kJ//mol` at `25^(@)C`
What is the value of `DeltaS^(@)` for the fuel cell at `25^(@)C`

Options

A. `0.41V`
B. `0.83V`
C. `-0.41V`
D. `-0.83V`

Answer

D

Explanation

The cell reaction is
`{:(2H_(2)+O_(2) hArr 2H_(2)O, E^(@) = 1.23 V, eq(i)),("For" O_(2)+3H_(2)O +4e^(-)hArr 4OH^(-),E^(@) =0.40V,eq.(ii)):}`
Hence for the reaction
`2H_(2)O +e^(-) hArr H_(2)+2OH^(-)` or `4H_(2)O +e^(-) hArr 2H_(2) +4OH^(-)` eq.(iii)
eq.(i)-eq(ii) gives the equation
`2H_(2)+4OH^(-) hArr 4H_(2)O +4e^(-)`
Hence for the equation
`4H_(2)O +4e^(-) hArr 2H_(2)+4OH^(-) E^(@) =- 0.83 V`
or for `2H_(2)O +2e^(-) hArr H_(2)+2OH^(-)`also `E^(@)` is `-0.83 V`
Hence choice (D) is correct and all others are incorrect.