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Same current is passed in series through a CuSO4 cell (Pt electrodes) and a dilute H2SO4 cell (inert electrodes). When 0.635 g of copper has been…

Numerical answer

Same current is passed in series through a

\mathrm{CuSO_4}

cell (Pt electrodes) and a dilute

\mathrm{H_2SO_4}

cell (inert electrodes). When

0.635\ \mathrm{g}

of copper has been deposited (

\mathrm{Cu}=63.5

), the total volume of gases (

\mathrm{H_2}

at the cathode plus

\mathrm{O_2}

at the anode) liberated in the acid cell at STP, in mL, is (

V_m=22.4\ \mathrm{L\,mol^{-1}}

)

Show answer and solution

Answer: 336

Charge passed:

0.635/63.5=0.0100

mol Cu, needing

2\times0.0100=0.0200

mol electrons. In the acid cell,

2\mathrm{H^+}+2e^-\to\mathrm{H_2}

gives

0.0100

mol

\mathrm{H_2}

(

224\ \mathrm{mL}

), and

2\mathrm{H_2O}\to\mathrm{O_2}+4\mathrm{H^+}+4e^-

gives

0.0200/4=0.00500

mol

\mathrm{O_2}

(

112\ \mathrm{mL}

). Total

=224+112=336\ \mathrm{mL}

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