Other Electrochemistry Concepts

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$Ag^{+}+e^{-}\rightarrow Ag$

$1 A \equiv1\frac{C}{S}$

$1:mol:e^{-}=96500:C$

A current is passed through a silver ion solution. How much silver is deposited after 10 minutes?

$Ag=107.9 \frac{g}{mol}$

Explanation

First we need to convert minutes to seconds (10min=600s), and knowing that $1A=1\frac{C}{s}$ , we can calculate the amount of silver deposited as follows:

$m_{Ag}=\frac{2.6: C}{s}\times600:s: \times\frac{1: mol: e^{-}}{96500: C}: \times\frac{1: molAg}{1: mol: e^{-}}: \times\frac{107.9: g: Ag}{1: molAg} =1.74:g:Ag$

If we follow the units carefully, we see that all of the units cancel out to give us a value in grams of silver, which is what we are looking for.

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