A) \[1.93\times {{10}^{4}}C\]
B) \[19.3\times {{10}^{5}}C\]
C) \[9.65\times {{10}^{4}}C\]
D) \[1.93\times {{10}^{5}}C\]
Correct Answer: D
Solution :
: \[{{H}_{2}}O\xrightarrow{{}}2{{H}^{+}}+\frac{1}{2}{{O}_{2}}+\underset{2F}{\mathop{2{{e}^{-}}}}\,\] Hence coulombs of electricity required for the oxidation of one mole of water to dioxygen \[=2\times 96500=193000\text{ }C=1.93\times {{10}^{5}}C\]
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