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The efficiency of a fuel cell is given b...

The efficiency of a fuel cell is given by:

A

`(DeltaS)/(DeltaG)`

B

`(DeltaR)/(DeltaG)`

C

`(DeltaG)/(DeltaS)`

D

`(DeltaG)/(DeltaH)`

Text Solution

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The correct Answer is:
D
Efficiency of a fuel cell `=(DeltaG)/(DeltaH)`
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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)

Thermodynamic efficiency of a cell is given by:

The thermodynamic efficiency of cell is given by

The thermodynamic efficiency of cell is given by

The efficiency of a fuel cell si 80% and the standard heat of reaction is -300kJ . The reaction involves two electrons in redox change. E^(c-) for the cell is

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