
In ${{H}_{2}}-{{O}_{2}}$ fuel cell, the reaction occurring at the cathode is:
(a) $2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)$
(b) ${{O}_{2}}(g)+2{{H}_{2}}O(l)+4{{e}^{-}}\to 4O{{H}^{-}}(aq)$
(c) ${{H}^{+}}+{{e}^{-}}\to \dfrac{1}{2}{{H}_{2}}$
(d) ${{H}^{+}}(aq)+O{{H}^{-}}(aq)\to {{H}_{2}}O(l)$
Answer
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Hint: The overall reaction in hydrogen-oxygen fuel is $2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)$. At the cathode, the electrons are used in the reaction while at the anode, the electrons are released.
Complete step by step answer:
We know that fuel cells are the devices that convert the energy produced during the combustion of fuels like hydrogen, methane, methanol, etc, directly into the electrical form of energy. A hydrogen-oxygen fuel cell is one such cell which has been very successfully used. On Apollo moon flight, this cell was used as the primary source of electrical energy.
The general ${{H}_{2}}-{{O}_{2}}$fuel cell consists of porous carbon electrodes containing suitable catalysts (generally finely divided platinum and palladium) incorporated in them. The solution used between the electrodes to act as the electrolyte is either concentrated $NaOH$$KOH$or. Into the $KOH/NaOH$solution, hydrogen and oxygen gases are bubbled through the porous electrodes.
The reaction occurring at anode, cathode, and overall reactions are given below:
At anode: $2{{H}_{2}}(g)+4O{{H}^{-}}(aq)\to 4{{H}_{2}}O(l)+4{{e}^{-}}$
At cathode: ${{O}_{2}}(g)+2{{H}_{2}}O(l)+4{{e}^{-}}\to 4O{{H}^{-}}(aq)$
And the overall reaction is$2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)$.
Thus, in these cells, the reactants are fed continuously to the electrodes, and the products are removed continuously from the electrolyte compartment.
These cells do not cause any pollution and are completely eco-friendly. They also have good efficiency and are higher than the thermal power plant. They never die because the continuous supply of the reactant is there.
Therefore, the correct answer is option (b) ${{O}_{2}}(g)+2{{H}_{2}}O(l)+4{{e}^{-}}\to 4O{{H}^{-}}(aq)$.
Note: The calculated efficiency of a hydrogen-oxygen fuel cell is 83%. There are some disadvantages of this fuel cell as the catalyst used is costly, all the three phases are used in this cell.
Complete step by step answer:
We know that fuel cells are the devices that convert the energy produced during the combustion of fuels like hydrogen, methane, methanol, etc, directly into the electrical form of energy. A hydrogen-oxygen fuel cell is one such cell which has been very successfully used. On Apollo moon flight, this cell was used as the primary source of electrical energy.
The general ${{H}_{2}}-{{O}_{2}}$fuel cell consists of porous carbon electrodes containing suitable catalysts (generally finely divided platinum and palladium) incorporated in them. The solution used between the electrodes to act as the electrolyte is either concentrated $NaOH$$KOH$or. Into the $KOH/NaOH$solution, hydrogen and oxygen gases are bubbled through the porous electrodes.
The reaction occurring at anode, cathode, and overall reactions are given below:
At anode: $2{{H}_{2}}(g)+4O{{H}^{-}}(aq)\to 4{{H}_{2}}O(l)+4{{e}^{-}}$
At cathode: ${{O}_{2}}(g)+2{{H}_{2}}O(l)+4{{e}^{-}}\to 4O{{H}^{-}}(aq)$
And the overall reaction is$2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l)$.
Thus, in these cells, the reactants are fed continuously to the electrodes, and the products are removed continuously from the electrolyte compartment.
These cells do not cause any pollution and are completely eco-friendly. They also have good efficiency and are higher than the thermal power plant. They never die because the continuous supply of the reactant is there.
Therefore, the correct answer is option (b) ${{O}_{2}}(g)+2{{H}_{2}}O(l)+4{{e}^{-}}\to 4O{{H}^{-}}(aq)$.
Note: The calculated efficiency of a hydrogen-oxygen fuel cell is 83%. There are some disadvantages of this fuel cell as the catalyst used is costly, all the three phases are used in this cell.
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