Redox Reactions & Electrochemistry Cheat Sheet

A printable reference covering oxidation numbers, balancing redox, galvanic cells, standard cell potential, the Nernst equation, and electrolysis for grades 11-12.

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Redox reactions involve electron transfer, which connects chemical change to electric current. This cheat sheet helps students identify oxidation and reduction, balance redox equations, and interpret electrochemical cells. It is especially useful for comparing galvanic cells, electrolytic cells, and reactions driven by voltage. Grade 11-12 students need these tools for stoichiometry, equilibrium, thermodynamics, and lab analysis. The main ideas are that oxidation is loss of electrons, reduction is gain of electrons, and electrons flow from the anode to the cathode through an external circuit. Standard cell potential is calculated with $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ , and spontaneity is connected to $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ . Nonstandard conditions are handled with the Nernst equation, $E = E^\circ - \frac{0.0592}{n}\log Q$ at $25^\circ\mathrm{C}$ . Electrolysis uses current to force nonspontaneous reactions, with charge given by $q = It$ and moles of electrons given by $n_{e^-} = \frac{q}{F}$ .

Key Facts

Oxidation is loss of electrons and reduction is gain of electrons, often remembered as $\mathrm{OIL\ RIG}$ .
The oxidizing agent is reduced, and the reducing agent is oxidized.
In any redox reaction, the total electrons lost must equal the total electrons gained, so $n_{\text{lost}} = n_{\text{gained}}$ .
For a galvanic cell, oxidation occurs at the anode, reduction occurs at the cathode, and electrons flow from anode to cathode.
Standard cell potential is calculated by $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ .
A reaction is spontaneous under standard conditions when $E^\circ_{\text{cell}} > 0$ and $\Delta G^\circ = -nFE^\circ_{\text{cell}} < 0$ .
At $25^\circ\mathrm{C}$ , the Nernst equation is $E = E^\circ - \frac{0.0592}{n}\log Q$ .
In electrolysis, charge is $q = It$ , and moles of electrons are found with $n_{e^-} = \frac{q}{F}$ , where $F = 96485\ \mathrm{C\ mol^{-1}}$ .

Vocabulary

Oxidation: Oxidation is the loss of electrons or an increase in oxidation number.
Reduction: Reduction is the gain of electrons or a decrease in oxidation number.
Anode: The anode is the electrode where oxidation occurs.
Cathode: The cathode is the electrode where reduction occurs.
Cell Potential: Cell potential is the voltage produced or required by an electrochemical cell, usually written as $E_{\text{cell}}$ .
Salt Bridge: A salt bridge allows ions to move between half-cells so charge does not build up.

Common Mistakes to Avoid

Confusing the oxidizing agent with the substance oxidized, because the oxidizing agent causes oxidation but is itself reduced.
Reversing anode and cathode labels, because oxidation always occurs at the anode and reduction always occurs at the cathode.
Changing coefficients before balancing atoms and charge in half-reactions, because redox balancing requires both mass balance and charge balance.
Using $E^\circ_{\text{anode}} - E^\circ_{\text{cathode}}$ , because standard cell potential must be calculated as $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ .
Forgetting to convert current and time into charge, because electrolysis calculations require $q = It$ before using $n_{e^-} = \frac{q}{F}$ .

Practice Questions

1 Calculate $E^\circ_{\text{cell}}$ for a galvanic cell with $E^\circ_{\text{cathode}} = +0.80\ \mathrm{V}$ and $E^\circ_{\text{anode}} = -0.76\ \mathrm{V}$ .
2 At $25^\circ\mathrm{C}$ , find $E$ for a cell with $E^\circ = 1.10\ \mathrm{V}$ , $n = 2$ , and $Q = 10.0$ using $E = E^\circ - \frac{0.0592}{n}\log Q$ .
3 How many moles of electrons pass through an electrolytic cell when $I = 2.50\ \mathrm{A}$ flows for $t = 1800\ \mathrm{s}$ , using $q = It$ and $F = 96485\ \mathrm{C\ mol^{-1}}$ ?
4 Explain why a salt bridge is necessary in a galvanic cell even though electrons travel through the wire.