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Application of Faraday's laws: An acidic solution of Cu^(2+) salt containing 0.4 g of Cu^(2+) is electroysed until all the copper is deposited. The electrolysis is futher continueed for seven more minutes with the volume of the solution kept at 100mL and the current at 1.0 amp. Calculate the volumes of gases evolved at NTP during the entire electrolysis. Strategy: Since we do not know which Cu^(2+) salt is electrolysed, we can say that (in the first stage of electrolysis)Cu^(2+) is reduced at the cathode to deposit Cu while H_(2)O is oxidized to liberate O_(g) at the anode. In the second stage of electrolysis, only water gets oxidized to librate O_(g) as well as reduced to librate H_(2)g |
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Answer» Solution :First stage of electrolysis: Anode (oxide): `2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-)` Cathode (red): `Cu^(2+)+2e^(-) rarr Cu` According to Faraday's laws Equivalents of `O_(2)` liberated = Euivalents of `Cu` deposited `= (0.4 g)/(31.8 g eq^(-1))` `{("Equivalent mass of an element"={("Atomic mass of element")/("VALENCY of element")}),(or),("Equivalent mass of metal"),(={("Atomic mass of metal")/("No of"e^(-)s " required to reduce the cation")}):}}` Since `1` equivalent of `O_(2)(g)` is `8 g (1//4 mol)`, the volume occupied by `1` equivalent of `O_(2)(g)` at `NTP` is `5.6 L`. Hence Volume of `O_(2)` (at `NTP`) liberated `= ((0.4)/(31.8)eq.) (5.6 L eq^(-1))` `= 0.07044 L` `= 70.44 mL` Second stage of Electrolysis: Anode (oxide): `2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq.)+4e^(-)` Cathode (red): `2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2OH^(-)(aq.)` According to Equation (3.34) `(m)/(E = (q)/(F) = (It)/(F)` THUS, EQS. Of `H_(2)(g)` liberated at cathode `=` Eqs. of `O_(2)(g)` liberatedet anode `=` Eqs. of ELECTRICITY passed `= ((1.2A)(7 xx 60 SEC))/(96,500 C eq^(-))` `(504)/(96,500)` Volume of `H_(2)(g)` evolved at `NTP` `= ((504)/(96,500)eq) (11.2 L eq^(-)) = 0.058492` `= 58.49 mL` Note that, `1` eq of `H_(2)` which is `1 g` (`1//2` mol), occupied `11.2 L` at `NTP` . Volume at `O_(2)(g)` evolved at `NTP` `((504)/(96,500)eq) (5.6 L eq^(-1))` ` 0.02924 L` `= 29.24 mL` Thus, during the entire process of electrolysis, Vol. of `H_(2)(g) = 58.49 mL` Vol. of `O_(2)(g) = (70.44 + 29.24)mL` `= 99.68 mL` |
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