InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
In the electrolysis of `7.2L` aqueous solution of `CuSO_(4)`, a current of `9.65A` passed for 2 hours. `a.` Calculate the weight of `Cu` deposited at cathode. `b.` Calculate the volume of `O_(2)` produced at anode at `27^(@)C` and `1 atm` pressure. `c.` Calculate the `pH` of the solution after electrolysis. `[` Atomic mass of `Cu=63g]` |
|
Answer» First method `(CuSO_(4) rarr Cu^(2+)+SO_(4)^(2-))` At cathode`:` `Cu^(2+)+2e^(-) rarr Cu(s)` At anode `:` `2H_(2)O rarr O_(2)+4H^(o+)+4e^(-)` `(` Since the oxidation potential of `H_(2)Ogt` oxidation potential of `SO_(4)^(2-)`, so oxidation of `H_(2)O` occurs at anode and `SO_(4)^(2-)` and `H^(o+)` ions remain in the solution to give an acidic solution `).` Number of faradays `=(Ixxt)/(96500C)=(9.65A xx 2 xx 3600 s )/(96500C mol^(-1))` `a.` Now from cathode `: (Cu^(2+)+2e^(-)rarr Cu)` `2e^(-) =2F=1` mol of `Cu=63 g ` of `Cu` `0.72F=(63g xx 0.72F)/(2)=22.68g` `b.` Now from anode `: (2H_(2)Orarr O_(2)+4H^(o+)+4e^(-))`. `4e^(-)=4-=1 mo l `of `o_(2)` `0.72F=(1xx0.72)/(4)=0.18 mol` Now using gas equation `: PV=nRT` `V=(nRT)/(P)=(0.18xx0.082xx300)/(1)=4.428L` of `O_(2)` `c.` Now from anode` (2H_(2)O rarr O_(2)+4H^(o+)+4e^(-))` `4e^(-)=4F=4 mol `of `H^(o+)` `[H^(o+)]=("Mol")/("Volume in L")=(0.72 mol)/(7.2L)=10^(-1)M` `:. pH=-log [10^(-1)]=1` `A.` At cathode`:` `1F=1` equivalent of `Cu=(63)/(2) `of `Cu` `0.7F=(63)/(2)xx0.72=22.68g` of `Cu` deposited `b.` At anode `:` `1F=1` equivalent of `O_(2)` `0.72F=0.72` equivalent of `O_(2)=(0.72)/(4)=0.18 mol ` of `O_(2)` `[{:(Since mol=("Equivalent")/("n factor")`or `("Equivalent")/("charge"),=,(0.72)/(4)),(2O^(-2) rarrO_(2)+4e^(-1)(n fact o r=4),=,):}]` Now for volume of `O_(2)`, process as in first method. `c." "1F=1` equivalent of `H^(o+)` ions. `0.72F=0.72` equivalent of `H^(o+)=0.72mol `of `H^(o+)` `[` Since` n` factor for `H^(o+)=1]` `:. [H^(o+)]=(0.72 mol)/(7.2L)=10^(-1)M` `:. pH=1` |
|