Fuel cells `:` Fuel cells are galvanic cells in which the chemical energy of fuel cell is directly converted into electrical energy. A type of fuel cell is a hydrogen `-` oxygen fuel cell. It consists of two electrodes made up of two porous graphite impregnated with a catalyst `(` platinum, silver, or metal oxide `).` The electrodes are placed in aqueous solution of `NaOH` . Oxygen and hydrogen are continuously fed into the cell. Hydrogen gets oxidized to `H^(o+)` which is neutralized by `overset(c-)(O)H, i.e.,` anodic reaction.

At cathode, `O_(2)` gets reduced to `overset(c-)(O)H`

Hence, the net reaction is
The overall reaction has
`DeltaH=-285.6 kJ mol^(-1)` and `DeltaG=-237.4 kJ mol^(-1)` at `25^(@)C`
If the cell voltage is `1.23V` for the `H_(2)-O_(2)` fuel cell and for the half cell `:`

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.