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Chemistry Question on Electrochemistry

How many Coulombs of electricity are required for the oxidation of one mol of water to dioxygen ?

A

9.65×104 C9.65 \times 10^4 \, C

B

1.93×104 C1.93 \times 10^4 \, C

C

1.93×105 C1.93 \times 10^5 \, C

D

19.3×105 C19.3 \times 10^5 \, C

Answer

1.93×105 C1.93 \times 10^5 \, C

Explanation

Solution

The electrode reaction for 1 mole H2OH _{2} O is

H2O→H2+12O2,O2−→12O2+2e−H _{2} O\to H _{2}+\frac{1}{2} O _{2}, O ^{2-} \to \frac{1}{2} O _{2}+2 e ^{-}
2H++2e−→H22 H ^{+}+2 e ^{-} \to H _{2}

or H2O→2H++12O2+2e−H _{2} O \to 2 H ^{+}+\frac{1}{2} O _{2}+2 e^{-}

Electricity required for oxidation of 1 mole of H2O=2F,1F=96500 CH _{2} O =2 F , 1 F =96500\, C
∴\therefore Total charge =2×96500=1.93×105 C=2 \times 96500=1.93 \times 10^{5} \,C