# Cathode and Anode Half-Cell Reactions Video

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• 0:04 Electrode Half-Reactions
• 2:47 Calculating Cell…
• 5:47 Corrosion
• 8:25 Lesson Summary
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Lesson Transcript
Instructor: Amy Meyers

Amy holds a Master of Science. She has taught science at the high school and college levels.

Learn how to write electrode half-reactions for cathodes and anodes. Discover how to calculate cell voltage potential when given a table of standard electrode potentials. Learn how to prevent corrosion using redox concepts and how to protect metal by cathodic protection.

## Electrode Half-Reactions

Batteries - they're useful things. Whether they're for your TV remote or your flashlight, it stinks when they lose power. In this lesson, you'll learn how batteries work and why they die.

You previously learned that an electrochemical cell consists of two half-cells: an anode and a cathode. The anode is the electrode where oxidation occurs and electrons are lost. The cathode is where reduction takes place and electrons are gained. These reactions don't occur separately; they must work together in a reduction-oxidation reaction.

For this next section, let's assume the electrodes are zinc and copper. There are other metals that can act as electrodes, but let's keep it simple and stick with these two. The half-reaction on the anode, where oxidation occurs, is Zn(s) = Zn2+ (aq) + (2e-). The zinc loses two electrons to form Zn2+. The half-reaction on the cathode where reduction occurs is Cu2+ (aq) + 2e- = Cu(s). Here, the copper ions gain electrons and become solid copper.

The entire reaction can be written by combining both half-reactions: Zn(s) + Cu2+ (aq) = Zn2+ (aq) + Cu(s). This equation for an electrochemical cell may also be shown as: Anode electrode - anode solution - cathode solution - cathode electrode, which would then look like: Zn(s) - Zn2+ (aq) - Cu2+ (aq) - Cu(s).

So, although both reactions happen at the same time, they are separated by a barrier. So, the only place for the electrons to move is through the wire connecting the anode and cathode. This is how you capture the electricity that is being produced.

Another type of electrochemical cell is a zinc-carbon dry cell. This is your size C and D batteries. These batteries have a zinc container that is the anode that is filled with a paste of MnO2, graphite and NH4 Cl. They have a carbon rod going down their middle, which is the cathode. When the circuit is closed, like when it is in a flashlight and the flashlight is turned on, the half-reaction on the anode is: Zn(s) = Zn2+ (aq) + 2e-. The reduction half-reaction on the cathode is: 2MnO2 (s) + H2 O (l) + 2e- = Mn2 O3 (s) + 2OH- (aq).

## Calculating Cell Voltage Potential

The electrochemical cells you have learned about are voltaic cells. A voltaic cell is one in which a spontaneous redox reaction produces electricity. In a voltaic cell, as you know, the oxidation reaction at the cathode pulls electrons into the wire that connects the cathode to the anode. It is called electric potential, and it is measured in volts (V). More specifically, electrode potential is the potential difference between an electrode and its solution, or the amount of pull on the electrons. It is the amount of energy needed to move an electric charge across a cell.

The two half-cells of the battery each have potential - potential to either give up electrons or take electrons. The difference between the two is the electrode potential of the whole cell, or voltage. Fortunately, the electrode potential isn't something you need to memorize. Scientists have determined standard electrode potentials for many different types of electrodes.

To determine the voltage of any cell, look on a table of electrode potentials. Subtract the electrode potential of the anode from that of the cathode, and you get the electrode potential of the cell, or voltage: Eo cell = Eo cathode - Eo anode.

This number is easy to calculate. For example, calculate the voltage of a cell for the reaction of liquid mercury electrode in a solution of mercury(I) nitrate and a cadmium metal electrode in a solution of cadmium nitrate. If the Eo isn't provided to you, you can refer to a chart for the value.

Electrode Electrode Reaction Eo Volts
Cd2+ Cd2+

The formula for the half-reactions are:

• Hg22+ + 2e- = 2Hg (l), which has an Eo = 0.850V
• Cd2+ (aq) + 2e- = Cd (s), which has an Eo = -0.403V
• Eo cell = Eo cathode - Eo anode

Eo cell = 0.850V - (-0.403V). Remember that two negatives make a positive. Eo cell = 1.253V.

Try another one.

• Zn2+ (aq) + 2e- = Zn (s), which has an Eo = -0.762V
• 2H2 O (l) + 2e- = H2 (g) + 2OH- (aq), which has an Eo = -0.828V

Eo cell = (-0.762 V) - (-0.828V). So, Eo cell = 0.066V.

## Corrosion

As you know, the Alaskan pipeline is a long steel pipe that brings oil from Alaska to the lower 48 states. What would happen if it corroded? Would oil spill all over our environment? How do we protect metal from corrosion?

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