A fuel cell is a cell that is continously supplied with an oxidant and a reductant so that if can deliver a current indefinitely.
Fuel cells offer the possibility of achieving high thermodynamic efficiency in the conversion of Gibbs energy into mechanical work.Internal combustion engines at best convert only the fraction `(T_2-T_1)//T_2` of the heat of combustion into mechanical work.
While the thermodynamic efficiency of the fuel cell is given by, `eta=(DeltaG)/(DeltaH)`, where `DeltaG` is the Gibbs energy change for the cell reaction and `DeltaH` is the enthalpy change of the cell reaction.A hydrogen-oxygen fuel cell may have an acidic or alkaline electrolyte.
`Pt|H_2(g)|H^(+)(aq.)||H_2O(l)|O_(2)(g)|Pt , (2.303 RT)/F=0.06`
The above fuel cell is used to produce constant current supply under constant temperature & 30 atm constant total pressure conditions in a cylinder.If 10 moles `H_2` and 5 moles of `O_2` were taken initially.
Rate of consumption of `O_2` is 10 milli moles per minute. The half-cell reactions are
`1/2O_2(g)+2H^+(aq)+2e^(-)toH_2O(l) E^(@)=1.246 V`
`2H^+(aq)+2e^(-) to H_2(g) E^(@)=0`
To maximize the power per unit mass of an electrochemical cell, the electronic and electrolytic resistances of the cell must be minimized.Since fused salts have lower electolytic resistances than aqueous solutions, high-temperature electrochemical cells are of special interest for practical applications.
Calculate e.m.f of the given cell at t=0.(log 2=0.3)

Options

A. `1.255 V`
B. `1.35 V`
C. `1.3 V`
D. `1.246 V`

Answer

C

Explanation

Cell reaction is, `H_2(g)+1/2O_2(g)toH_2O(l)`
`E_(Cell)=1.246-0.06/2"log"1/(P_(H_(2))xxsqrtP_(O_(2)))=1.246-0.06/2"log"1/(20sqrt10)`
`=1.246+0.06/2(1.3+1/2)=1.246+0.054=1.3 V`