What mass of copper is deposited at the cathode when a steady current of 0.50\text{ A} is … — JAMB UTME

What mass of copper is deposited at the cathode when a steady current of

0.50\text{ A}

is passed through an aqueous solution of copper(II) tetraoxosulfate(VI) for

1930\text{ s}

? [

\text{Molar mass of Cu} = 64\text{ g mol}^{-1}, 1\text{ F} = 96,500\text{ C mol}^{-1}

]

The correct mass of copper deposited is

0.32\text{ g}

Passing

0.50\text{ A}

for

1930\text{ s}

transfers

965\text{ C}

of electricity, which equals

0.01\text{ mol}

of electrons. Because reduction of

Cu^{2+}

requires

2

moles of electrons per mole of copper metal deposited (

Cu^{2+} + 2e^- \rightarrow Cu

0.005\text{ mol}

of copper is produced. Multiplying

0.005\text{ mol}

by the molar mass of copper (

64\text{ g mol}^{-1}

) gives

0.32\text{ g}

Calculate the total quantity of electricity (

Q

) transferred in Coulombs.

Q = I \times t = 0.50\text{ A} \times 1930\text{ s} = 965\text{ C}

Electric charge is the product of current in amperes and time in seconds.

Determine the number of moles of electrons transferred.

\text{Moles of } e^- = \frac{965\text{ C}}{96,500\text{ C mol}^{-1}} = 0.01\text{ mol}

One Faraday (

96,500\text{ C}

) corresponds to one mole of electrons.

Use the cathode half-reaction stoichiometry to find the moles of copper deposited.

Cu^{2+} + 2e^- \rightarrow Cu(s) \implies \text{Moles of Cu} = \frac{0.01\text{ mol } e^-}{2} = 0.005\text{ mol}

Copper(II) ions require 2 moles of electrons per mole of copper metal deposited.

Calculate the mass of copper deposited.

\text{Mass} = \text{moles} \times \text{molar mass} = 0.005\text{ mol} \times 64\text{ g mol}^{-1} = 0.32\text{ g}

Mass is found by multiplying the quantity in moles by the molar mass.

Faraday's First Law of Electrolysis and Quantitative Stoichiometry of Electrode Reactions

Question