Question

A constant current was passed through a solution of AuCl-4 ions between gold electrodes.After a period of 10.00 minutes the cathode increased in weight by 1.314grams.how much charge was passed and what was the current I?

Answer 1

Question #63709, Chemistry / Other

A constant current was passed through a solution of $\mathrm{AuCl}_4^-$ ions between gold electrodes. After a period of 10.00 minutes the cathode increased in weight by 1.314 grams. How much charge was passed and what was the current I?

Solution:

Faraday's laws can be summarized by

$$m = \frac{Q M}{F z}$$ $$Q = I t$$

- $m$ is the mass of the substance liberated at an electrode in grams

- $Q$ is the total electric charge passed through the substance in coulombs

- $F = 96485\ \mathrm{C\cdot mol^{-1}}$ is the Faraday constant

- $M$ is the molar mass of the substance in grams per mol

- $z$ is the valency number of ions of the substance (electrons transferred per ion).

$$Q = \frac{m F z}{M}$$

$\mathrm{AuCl}_4^-$ ions – $\mathrm{Au}^{+3}$ reduces to $\mathrm{Au}^0$. $z = 3$.

$\mathrm{M(Au)} = 196.967\ \mathrm{g/mol}$

$$Q = \frac{1.314\ \mathrm{g} \times 96485\ \mathrm{C/mol} \times 3}{196.967\ \mathrm{g/mol}} = 1931.0\ \mathrm{C}$$ $$I = \frac{Q}{t} = \frac{1931.0}{600\ \mathrm{s}} = 3.22\ \mathrm{A}$$

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