Electrolysis involves electronation and de-electronation at the respective electodes. Anode electrolytic cell is the electode at which de-electronation takes place whereas at cathod electronation is noticed. If two or more ions of same charge are to be electonated or deelectronate the ion having lasser discharge potential is discharged. Discharge potential of anion refers for ` EP_(OP)^(@) ` of `E_("RP")^(@)` as the case may be. The products formed at either electode is given in terms
Faraday's laws of electrolusis i.e.` w=(Eit)/(96500)`
During electrolysis of `CuSO_(4(aq))` , the pH of solution becomes

Options

A. `322.56 L `
B. `32. 256 L `
C. `3.22 mL `
D. ` 1.612 xx 10^(2) mL `

Answer

D

Explanation

`96500 CACl to (10)/(4)" mole " of O_2 , 9.65 xx 60 xx 60 CaCl to n_(O_2) `
`n_(O_2) =(9.65 xx 60 xx 60 xx (1)/(4))/(9.6500) = 0.09 ` mole , `C_(8) H_(8) + 25O_2 to 8CO_2 + 9H_2O`
` 1 xx 22.4 " lit to 25 mole of "O_2 , V =? To 0.09 " mole of " O_2`
` V=(22.4 xx 00.9)/((25)/(2)) = 0.19128 "lit" = 161.28 mL = 1.61 xx 10^(2) ml`