Question

Which of the following reactions is used to make a fuel cell?
(A) \[2{{\rm{H}}_{\rm{2}}}\left( g \right) + {{\rm{O}}_{\rm{2}}}\left( g \right) \to 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\]
(B) \[2{\rm{Fe}}\left( s \right) + {{\rm{O}}_{\rm{2}}}\left( g \right) + 4{{\rm{H}}^ + }\left( {aq} \right) \to 2{\rm{F}}{{\rm{e}}^{2 + }}\left( {aq} \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\]
(C) \[{\rm{Pb}}\left( {\rm{s}} \right) + {\rm{Pb}}{{\rm{O}}_{\rm{2}}}\left( s \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\left( {aq} \right) \to 2{\rm{PbS}}{{\rm{O}}_{\rm{4}}}\left( s \right) + {{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\]
(D) \[{\rm{Cd}}\left( s \right) + 2{\rm{Ni}}{\left( {{\rm{OH}}} \right)_3}\left( s \right) \to {\rm{CdO}}\left( s \right) + 2{\rm{Ni}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\]

Answer 1

Hint: A fuel cell belongs to the class of electrochemical cells which do the work of the generation of electrical energy from fuel by an electrochemical equation. For the generation of electrical energy, they require supply of fuel in a continuous manner and an oxidising agent, mostly Oxygen. Some of the fuels used in the fuel cell are carbon-based fuel such as biogas and natural gas.

Complete Step by Step Solution:
Let’s understand the fuel cell in detail. A fuel cell has a similar setup of electrochemical cells. A fuel cell consists of an anode, cathode and electrolyte. The role of an electrolyte is in the motion of protons.

Now, we will discuss the working of a fuel cell. The reaction of hydrogen and oxygen can generate electricity via a fuel. In the fuel cell, there is the transfer of oxygen and hydrogen via carbon electrodes into a solution of sodium hydroxide (concentrated form). The reactions at the cathodes and anodes take place as given below.

Reaction at cathode: \[{{\rm{O}}_{\rm{2}}} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + 4{e^ - } \to 4{\rm{O}}{{\rm{H}}^ - }\]
Reaction at anode: \[2{{\rm{H}}_{\rm{2}}} + 4{\rm{O}}{{\rm{H}}^ - } \to 4{{\rm{H}}_{\rm{2}}}{\rm{O}} + 4{e^ - }\]

The net cell reaction can be written as follows:
\[2{{\rm{H}}_{\rm{2}}} + {{\rm{O}}_{\rm{2}}} \to 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\]
Therefore, the net equation for the fuel cell is, \[2{{\rm{H}}_{\rm{2}}}\left( g \right) + {{\rm{O}}_{\rm{2}}}\left( g \right) \to 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)\].
Hence, option A is right.

Note: It is to be noted that the fuel cell reaction is a very slow process. To make the reaction take place at a faster rate, finely divided catalysts namely palladium, and platinum are used. These catalysts are used in the form of finely divided to increase effective surface area.