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Chemistry Question on Electrochemistry

In H2−O2H_2 - O_2 fuel cell the reaction occurring at cathode is

A

2H2(g)+O2(g)→2H2O(l)2H_{2} (g) + O_2 (g) \to 2H_2O(l)

B

O2(g)+2H2O(l)+4e−→4O−H(aq)O_2 (g)+2H_2O(l)+4e^{-} \to 4 \overset{-}{O}H (aq)

C

H++e−→12H2H^{+} +e^{-} \to \frac{1}{2} H_{2}

D

H−(aq)+O−H(aq)→H2O(l)H^{-}(aq)+O^{-}H (aq) \to H_2 O(l)

Answer

O2(g)+2H2O(l)+4e−→4O−H(aq)O_2 (g)+2H_2O(l)+4e^{-} \to 4 \overset{-}{O}H (aq)

Explanation

Solution

In a fuel cell,
Cathode reaction can be written as:
O2(g)+2H2O(l)+4e−→4OH−(aq)O _{2}( g )+2 H _{2} O ( l )+4 e ^{-} \rightarrow 4 OH ^{-}( aq )
Anode reaction can be written as:
H2+2OH−→2H2O+2e−H _{2}+2 OH ^{-} \rightarrow 2 H _{2} O +2 e ^{-}
Net cell reaction:
2H2+O2→2H2O2 H _{2}+ O _{2} \rightarrow 2 H _{2} O