Question

1mole of electron is passed through three electrolytic cells in series. First contain silver ions the second contain zinc ions and third contain ferricions assuming that the only cathode reaction in each cells reduces the ions of metal .

Calculate the number of mole of each metal deposited?

Answer 1

In an electrolytic cell, 1 mole of electrons will deposit 1 mole of silver, 0.5 moles of zinc, and 1/3 moles of iron from their respective ions Ag+, Zn2+, and Fe3+, according to the stoichiometry of each reduction reaction at the cathode.

Step-by-step explanation:

When 1 mole of electrons is passed through a series of electrolytic cells, the amount of metal deposited depends on the stoichiometry of the reduction reaction at the cathode.

For silver ions (Ag+), the cathode reaction is:

• Ag+ (aq) + e- → Ag(s)

Thus, 1 mole of electrons will reduce 1 mole of Ag+ ions to 1 mole of solid silver metal.

For zinc ions (Zn2+), the cathode reaction is:

• Zn2+ (aq) + 2e- → Zn(s)

Here, it takes 2 moles of electrons to reduce 1 mole of Zn2+ ions, so 1 mole of electrons will reduce only 0.5 moles of Zn2+ ions to solid zinc metal.

Lastly, for ferric ions (Fe3+), the cathode reaction is:

• Fe3+ (aq) + 3e- → Fe(s)

In this case, 3 moles of electrons are required to reduce 1 mole of Fe3+, hence 1 mole of electrons will reduce only 1/3 moles of Fe3+ ions to solid iron metal.