Electrochemistry: How much zinc is consumed and how much copper is deposited if a current of 1.0 A flows for 1.0 hour?

Question

How much zinc is consumed and how much copper is deposited if a current of 1.0 A flows for 1.0 hour?

Answer 1

Let us first look at the redox reactions:

Zn(s) ===> Zn²⁺ + 2e^-

Cu²⁺(aq) + 2e^- ===> Cu(s)

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Zn(s) + Cu²⁺(aq) ===> Zn²⁺(aq) + Cu(s)

Now, recall that 1 amp (1 A) = 1 coulomb/sec (1 C/s)

1.0 A for 1hr = 1.0 A for 3600 seconds = 3600 C

It turns out that it takes 96,485 coulombs (1 Faraday) to equate to 1 mole of electrons.

So, now we can find the moles of electrons transferred in the redox reaction above:

3600 C x 1 mole e-/96485 C = 0.0373 moles e-

But it takes 2 moles of e- for each mole of Zn consumed and for each mole of Cu deposited. Thus ...

0.0373 moles e- x 1 mole Zn/2 moles e- = 0.0187 moles of Zn consumed = moles Cu deposited.

If you actually want the mass of each, then proceed as follows:

mass of Zn consumed = 0.0187 moles Zn x 65.4g/mol = 1.22 g Zn = 1.2 g Zn (to 2 significant figures)

mass of Cu deposited = 0.0187 moles Cu x 63.5 g/mol = 1.19 g Cu = 1.2 g Cu (to 2 significant figures)