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

Question

The efficiency of a fuel cell is 80 % and the standard the heat of reaction is -300 kJ. The reaction involves two electrons in redox change. E ominus for the cell is (A) 1.24 V (B) 2.48 V (C) 0 V (D) 0.62 V

Tardigrade Admin · Accepted Answer

Efficiency =(Δ G ominus/Δ H ominus)=-(n E ominusF /Δ H)=80 E ominus=(80 ×(-300) × 103/2 × 96500 × 100)=1.24 V

Q. The efficiency of a fuel cell is $80%$ and the standard the heat of reaction is $- 300 k J$ . The reaction involves two electrons in redox change. $E^{⊖}$ for the cell is

$1.24 V$

$2.48 V$

$0 V$

$0.62 V$

Efficiency $= \frac{Δ G ^{⊖}}{Δ H ^{⊖}} = - \frac{n E ^{⊖} F}{Δ H} = 80$
$E^{⊖} = \frac{80 \times ( - 300 ) \times 1 0 ^{3}}{2 \times 96500 \times 100} = 1.24 V$

Q. The efficiency of a fuel cell is 80% and the standard the heat of reaction is −300kJ. The reaction involves two electrons in redox change. E⊖ for the cell is

1.24V

2.48V

0V

0.62V