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Chemistry Question on Electrochemistry

Anode reaction of a fuel cell is

A

Zn(Hg)+2OH−→ZnO(s)+H2O+2e−Zn ( Hg )+2 OH ^{-} \rightarrow ZnO _{(s)}+ H _{2} O +2 e^{-}

B

Pb(s)+SO4(ag.)2−→PbSO4(s)+2e−Pb (s)+ SO _{4( ag .)}^{2-} \rightarrow PbSO _{4(s)}+2 e^{-}

C

2H2(g)+4OH−(aq)→4H2O(l)+4e−2 H _{2}(g)+4 OH ^{-}{ }_{(a q)} \rightarrow 4 H _{2} O _{(l)}+4 e^{-}

D

2Fe(s)→2Fe2++4e−2 F e(s) \rightarrow 2 F e^{2+}+4 e^{-}

Answer

2H2(g)+4OH−(aq)→4H2O(l)+4e−2 H _{2}(g)+4 OH ^{-}{ }_{(a q)} \rightarrow 4 H _{2} O _{(l)}+4 e^{-}

Explanation

Solution

Anode reaction for a fuel cell (cell that converts the chemical energy of a fuel directly into electrical energy) is
2H2(g)+4OH−(aq)→4H2O(l)+4e−2 H _{2}(g)+4 OH ^{-}(a q) \rightarrow 4 H _{2} O (l)+4 e^{-}