Electrolusis involves electronation and de-electronation at the redpective electrodes. Anode of electrolytic cell is the electrode at which de-electronation takes place whereas at cathode electronation is noticed. If two or more ions of same charge are to be electronated or de-electronated, the ion having lesser discharge potential is dischargeed. Discharge potential of an ion refers for `E_(OP)^(@)` or `E_(RP)^(@)` as the case may be. The products formed at either electrode is given in terms of Faraday's laws of electrolysis i.e., `w = (Eit)/(96500)`
F. The volume of octane required to be used for its combustion by the oxygen liberated during electrolysis of an `NaNO_(3(aq.))` by passing `9.65` ampere current for `1` hr. is :

Options

A. 322.56 mL
B. 32.256 mL
C. 3.22 mL
D. `1.612xx10^(2) mL`

Answer

D

Explanation

Anode : `2OH^(-) rarr H_(2)O + 1//2O_(2) + 2e`,
Cathode : `2H^(+) + 2e rarr H_(2)`
Eq. of `O_(2)` formed `= (9.65 xx 10 xx 60 xx 60)/(96500) = 0.36`
`:.` Mole of `O_(2)` formed `= (0.36)/(4) = 0.09`
`C_(8)H_(18) + (25)/(2)O_(2) rarr 8CO_(2) + 9H_(2)O`
`:.` Mole of `C_(8)H_(18) = (2)/(25) xx 0.09 = 7.2 xx 10^(-3)`
`:. V_(C_(8)H_(18)) = 7.2 xx 10^(-3) xx 22400 = 161.2 mL`