Electrical current is defined as the rate of flow for any charged species. Most relevant to this discussion is the flow of electrons. Current is measured in a composite unit called an ampere, defined as one coulomb per second (A = 1 C/s). The charge transferred, *Q*, by passage of a constant current, *I*, over a specified time interval, *t*, is then given by the simple mathematical product

$$Q=It$$

When electrons are transferred during a redox process, the stoichiometry of the reaction may be used to derive the total amount of (electronic) charge involved. For example, the generic reduction process

$$M^{n+}(aq) + ne^- ⟶ M(s)$$

involves the transfer of *n* mole of electrons. The charge transferred is, therefore,

$$Q=nF$$

where *F* is Faraday’s constant, the charge in coulombs for one mole of electrons. If the reaction takes place in an electrochemical cell, the current flow is conveniently measured, and it may be used to assist in stoichiometric calculations related to the cell reaction.

**Converting Current to Moles of Electrons**

In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?

**Solution**

Faraday’s constant can be used to convert the charge (*Q*) into moles of electrons (*n*). The charge is the current (*I*) multiplied by the time

From the problem, the solution contains AgNO_{3}, so the reaction at the cathode involves 1 mole of electrons for each mole of silver

$$\text{cathode: }Ag^+(aq) + e^− ⟶ Ag(s)$$

The atomic mass of silver is 107.9 g/mol, so

$$\text{mass Ag}=0.3817\;mol\;e^− × \frac{1\;mol\;Ag}{1\;mol\;e^−} × \frac{107.9\;g\;Ag}{1\;mol\;Ag} = 41.19\;g\;Ag$$

**Check Your Learning**Aluminum metal can be made from aluminum(III) ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 25.0 A passed through the solution for 15.0 minutes?

## Answer:

$Al^{3+}(aq)+3e^−⟶Al(s)$; 0.0777 mol Al = 2.10 g Al.

**Time Required for Deposition**

In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m^{2} from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm^{3}.

**Solution**

First, compute the volume of chromium that must be produced (equal to the product of surface area and thickness):

Use the computed volume and the provided density to calculate the molar amount of chromium required:

$$\text{mass}=\text{volume}\times\text{density}=33\;\require{enclose}\enclose{horizontalstrike}{cm^3}\times\frac{7.19\;g}{\enclose{horizontalstrike}{cm^3}}=237\;g\;Cr$$ $$mol\;Cr=237\;g\;Cr\times\frac{1\;mol\;Cr}{52.00\;g\;Cr}=4.56\;mol\;Cr$$The stoichiometry of the chromium(III) reduction process requires three moles of electrons for each mole of chromium(0) produced, and so the total charge required is:

$$Q=4.56\;mol\;Cr\times\frac{3\;mol\;e^-}{1\;mol\;Cr}\times\frac{96485\;C}{mol\;e^-}=1.32×10^6\;C$$Finally, if this charge is passed at a rate of 33.46 C/s, the required time is:

$$t=\frac{Q}{I}=\frac{1.32×10^6\;C}{33.46\;C/s}=3.95×10^4\;s=11.0\;hr$$**Check Your Learning**

What mass of zinc is required to galvanize the top of a 3.00 m × 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO_{3})_{2} and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm^{3}.

## Answer:

11.8 kg Zn requires 382 hours.