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Ion Electron Method

The electrode...

Question

The electrode reaction that takes place at the anode of $C H_{4} - O_{2}$ fuel cell is:

2 O_{2} + 8 H^{+} + 8 e^{-} \to 4 H_{2} O

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C H_{4} + 2 H_{2} O \to C O_{2} + 8 H^{+} + 8 e^{-}

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C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O

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2 H^{+} + 2 e^{-} \to H_{2}

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Solution

The correct option is C $C H_{4} + 2 H_{2} O \to C O_{2} + 8 H^{+} + 8 e^{-}$
The cathode reactions in methane-oxygen fuel cells is
Anode reaction: $C H_{4} + 2 H_{2} O ⟶ C O_{2} + 8 e^{-} + 8 H^{+}$
Cathode reaction: $8 H^{+} + 2 O_{2} + 8 e^{-} ⟶ 4 H_{2} O$
Overall reaction: $C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$ .

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