What mass of silver is deposited during electrolysis when a current of 0.8 A flows for 25 minutes?

Answer Details

Faraday's first law of electrolysis states that the mass deposited at an electrode is proportional to the quantity of charge passed, \(m = ZQ = ZIt\), where \(Z\) is the electrochemical equivalent of the substance. The whole calculation therefore begins with the charge.

Convert the time to seconds first, since the ampere is a coulomb per second:

\[t = 25\times 60 = 1500\ \text{s},\qquad Q = It = 0.8\times 1500 = 1200\ \text{C}.\]

For silver, one mole of \(\text{Ag}^{+}\) ions carries one faraday of charge, so depositing \(108\ \text{g}\) requires \(96\,500\ \text{C}\). This gives

\[Z_{\text{Ag}} = \frac{108}{96\,500} = 1.118\times 10^{-3}\ \text{g C}^{-1},\]

and hence

\[m = Z_{\text{Ag}}\,Q = 1.118\times 10^{-3}\times 1200 = 1.34\ \text{g}.\]

The mass of silver deposited is about \(1.34\ \text{g}\).

The most frequent error is leaving the time in minutes, which makes the charge \(20\ \text{C}\) and the mass a hundredth of the true value, landing near the small figures offered here. A second error is dividing by a valency of \(2\); silver is monovalent, unlike copper in \(\text{Cu}^{2+}\), so no factor of two appears. Remember the routine: seconds, then coulombs, then multiply by the electrochemical equivalent.