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A current of 10A was allowed to pass thr...

A current of 10A was allowed to pass through acidulated water for 3 min 13s. What mass of water will be decomposed?

Text Solution

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Quantity of electricity `=10xx(3xx60+13)=1930C`
Cathode reaction : `2H^(+)(aq)+2e rarr H_(2)(g)`,
Anode reaction : `H_(2)O(l)rarr 2H^(+)(aq)+(1)/(2)O_(2)(g)+2e` and
Overall reaction : `H_(2)O(l)rarr H_(g)+(1)/(2)O_(2)(g)`.
Hence, the amount of electricity required to produce 1 mol of `H_(2)` (or, 0.5 mol of `O_(2)`)=`2xx96500C.`
`therefore" 1930 C of electricity will produce "(1xx1930)/(2xx96500)="0.01 mol of "H_(2)" gas. "`
From the overall reaction we see that 0.01 mol of `H_(2)-=0.01` mol of `H_(2)O-=0.18g` of water. Therefore, the amount of water that will be decomposed is 0.18 g.
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