Fuel cells are galvanic cells in which the chemical energy of fuel is directly converted into electrical energy. 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^(oplus)` which is neutralized by `overset(Θ)(OH)`, ie., anodic reaction.
`H_(2) hArr 2H^(oplus) + 2e^(-)`
`2H^(oplus) + 2 overset(Θ)(OH) hArr 2H_(2)O`
`H_(2) + 2 overset(Θ)(OH) hArr 2H_(2)O + 2e^(-)`
At cathode, `O_(2)` gets reduced to `overset(Θ)(OH)` i.e., `O_(2) + 2H_(2)O + 4e^(-) hArr overset(Θ)(4)OH`
Hence, the net reaction is `2H_(2) + O_(2) hArr 2H_(2)O`
The overall reaction has `Delta H= -285.6kJ mol^(-1) and Delta G= -237.4kJ 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: `O_(2) + 2H_(2) O + 4e^(-) hArr 4 overset(Θ)(OH) " has " E^(Θ) = 0.40V`, then `E^(Θ) " for " 2H_(2)O + 2e^(-) hArr H_(2) + 2 overset(Θ)(OH)` will be

Options

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

Answer

D

Explanation

Cathode : `O_(2) + 2H_(3)O + 4e^(-) rarr overset(Θ)(4OH)`
`ul("Anode" H_(2) + overset(Θ)(2OH) rarr 2H_(2)O + 2e^(-))`
`ul("Cell reaction" O_(2)+ 2H_(2) hArr 2H_(2)O)`
`rArr E_("cell")^(Θ) =1.23= 0.4- (E_("red")^(Θ))_(a) rArr (E_("red")^(Θ))_(a)= -0.83V`