A hydrogen-oxygen fuel cell operates at a terminal voltage of $0.72 V$ . The electrical work obtained when $2.0 mol$ of $H X_{2}$ is consumed is ( $F = 96500 C mo l^{- 1}$ )

Answer: C

Each

H X_{2}

gives

2

electrons (

H X_{2} 2 H X^{+} + 2 e X^{-}

H X_{2} + 2 OH X^{-} 2 H X_{2} O + 2 e X^{-}

), so

2.0 mol

H X_{2}

passes

4.0 mol

of electrons. Work

= n FE = 4.0 \times 96500 \times 0.72 = 2.78 \times 1 0^{5} J = 278 kJ

. Option (B) forgets that each

H X_{2}

gives two electrons.