The electrolyte used in $H_2 - O_2$ fuel cell is ________.

Question

Consider following reaction :
$Fe(OH)_2 + 2e^- \rightarrow Fe(s) + 2OH^-$ , $E^{\circ} = -0.88 \text{ V}$
$AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$ , $E^{\circ} = 0.07 \text{ V}$
Select the correct statement.
(1) $E^{\circ}_{cell} = -0.95 \text{V}$
(2) $E^{\circ}_{cell}$ is an extensive property
(3) $Fe$ is getting reduced
(4) Net cell reaction: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2 + 2Ag(s) + 2Br^-(aq)$

Answer 1

Step 1: Understanding the Concept:
A fuel cell is an electrochemical cell that converts the chemical energy of a fuel (hydrogen) and an oxidizing agent (oxygen) into electricity through a pair of redox reactions.
Step 3: Detailed Explanation:
In a standard alkaline $H_2 - O_2$ fuel cell, the electrolyte used is a concentrated aqueous solution of potassium hydroxide ($KOH$) or sodium hydroxide ($NaOH$).
Aqueous $KOH$ is preferred because of its high ionic conductivity.
The electrodes are made of porous carbon impregnated with catalysts like platinum.
The hydroxide ions ($OH^-$) migrate through the electrolyte to complete the circuit.
Step 4: Final Answer:
The correct electrolyte is aqueous $KOH$.

The electrolyte used in $H_2 - O_2$ fuel cell is ________.

Show Hint

The Correct Option is B

Solution and Explanation

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