Answer to Question #156147 in General Chemistry for RAHUL

Question

In a H2 fuel cell, electricity is produced by oxidation-reduction reaction between

H2 and O2 gases. Draw a diagram of H2 – O2 fuel cell and write down the cell

reaction of that fuel cell. At STP 89.6 L H2 gas is passed in H2 fuel cell which

reacts with O2 for 20 minutes. What is the average strength of electricity (in

ampere) produced? If total current produced is passed in CuSO4 solution of

electrolytic cell, then what amount of Cu metal will be deposited at cathode?

Answer 1

The redox changes in fuel cell are

$2H_{2(g) }+ O_{2(g)} \to 2H_2O_{(l)}$

At anode:

$H_{2(g)}+2OH^- \to2H_2O_{(l)} + 2e^-$

At cathode:

$O_{2(g)} +2H_2O_{(l)} +4e^ - \to 4OH^-$

Therefore, moles of H₂ reacting = $\dfrac{89.6}{22.4} = 4\ moles$

Therefore, equivalent of H₂ used = 4 moles also

Now, $\dfrac{w}{E}=\dfrac{i×t}{96500}$

Therefore, $8 = \dfrac{i×20×60}{96500}$

i = 643.33 A

eq. moles of H₂ formed = eq. moles of Cu formed = 8 moles

Hence, $W_{Cu} = \dfrac{8×63.5}{2}$ = 254g