Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

Standard Reduction Potentials Reference cheat sheet - grade 11-12

Click image to open full size

Standard reduction potentials help students predict how electrons move in electrochemical cells. This reference explains how to read a reduction potential table, identify the cathode and anode, and calculate standard cell voltage. It is useful for solving galvanic cell problems, ranking oxidizing agents, and connecting voltage to spontaneity.

The core idea is that each half-reaction is written as a reduction and assigned an EE^\circ value under standard conditions. The more positive reduction potential occurs at the cathode, while the half-reaction with the lower reduction potential is reversed for oxidation at the anode. Important formulas include Ecell=EcathodeEanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}, ΔG=nFEcell\Delta G^\circ = -nFE^\circ_{\text{cell}}, and the Nernst equation for nonstandard conditions.

Key Facts

  • Standard reduction potentials, EE^\circ, are measured in volts under standard conditions of 1M1\,\text{M} solutions, 1atm1\,\text{atm} gases, and 25C25^\circ\text{C}.
  • The standard hydrogen electrode is the reference half-cell and is assigned E=0.00VE^\circ = 0.00\,\text{V} for 2H++2eH22\text{H}^+ + 2e^- \rightarrow \text{H}_2.
  • For a galvanic cell, the half-reaction with the more positive EE^\circ is reduced at the cathode.
  • When using a standard reduction potential table, calculate cell voltage with Ecell=EcathodeEanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}.
  • If a reduction half-reaction is reversed to show oxidation, the sign of its EE^\circ value changes.
  • Do not multiply EE^\circ values by coefficients because voltage is an intensive property.
  • A reaction is spontaneous under standard conditions when Ecell>0E^\circ_{\text{cell}} > 0 and ΔG<0\Delta G^\circ < 0.
  • Free energy and voltage are related by ΔG=nFEcell\Delta G^\circ = -nFE^\circ_{\text{cell}}, where F=96485Cmol1F = 96485\,\text{C}\,\text{mol}^{-1}.

Vocabulary

Standard reduction potential
The voltage of a reduction half-reaction measured against the standard hydrogen electrode under standard conditions.
Cathode
The electrode where reduction occurs and electrons are gained by a chemical species.
Anode
The electrode where oxidation occurs and electrons are lost by a chemical species.
Oxidizing agent
A substance that causes another substance to be oxidized by accepting electrons and being reduced.
Reducing agent
A substance that causes another substance to be reduced by donating electrons and being oxidized.
Nernst equation
An equation that adjusts cell voltage for nonstandard concentrations using E=E0.0592nlogQE = E^\circ - \frac{0.0592}{n}\log Q at 25C25^\circ\text{C}.

Common Mistakes to Avoid

  • Adding reduction potentials instead of using Ecell=EcathodeEanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} is wrong because the table lists both half-reactions as reductions.
  • Multiplying EE^\circ by reaction coefficients is wrong because voltage does not depend on the amount of substance reacting.
  • Forgetting to reverse the sign when a half-reaction is written as oxidation is wrong because the listed value applies only to the reduction direction.
  • Choosing the anode as the half-reaction with the larger EE^\circ is wrong in a galvanic cell because the larger reduction potential is reduced at the cathode.
  • Using the Nernst equation without balancing electrons is wrong because nn must equal the number of electrons transferred in the balanced redox reaction.

Practice Questions

  1. 1 Given E(Cu2++2eCu)=+0.34VE^\circ(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}) = +0.34\,\text{V} and E(Zn2++2eZn)=0.76VE^\circ(\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}) = -0.76\,\text{V}, calculate EcellE^\circ_{\text{cell}} for a zinc-copper galvanic cell.
  2. 2 Given E(Ag++eAg)=+0.80VE^\circ(\text{Ag}^+ + e^- \rightarrow \text{Ag}) = +0.80\,\text{V} and E(Fe2++2eFe)=0.44VE^\circ(\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}) = -0.44\,\text{V}, identify the cathode and calculate EcellE^\circ_{\text{cell}}.
  3. 3 For a cell with Ecell=+1.10VE^\circ_{\text{cell}} = +1.10\,\text{V} and n=2n = 2, calculate ΔG\Delta G^\circ using ΔG=nFEcell\Delta G^\circ = -nFE^\circ_{\text{cell}} and F=96485Cmol1F = 96485\,\text{C}\,\text{mol}^{-1}.
  4. 4 Explain why Ag+\text{Ag}^+ is a stronger oxidizing agent than Zn2+\text{Zn}^{2+} when E(Ag+/Ag)=+0.80VE^\circ(\text{Ag}^+ / \text{Ag}) = +0.80\,\text{V} and E(Zn2+/Zn)=0.76VE^\circ(\text{Zn}^{2+} / \text{Zn}) = -0.76\,\text{V}.

Understanding Standard Reduction Potentials Reference

A reduction potential is not a property of one metal strip by itself. It describes a half-cell connected to a reference system. The standard hydrogen electrode gives chemists a shared zero point, much like sea level gives a shared height reference.

A table therefore compares the tendency of many substances to gain electrons. Species near the top of a typical table are strong oxidizing agents because they readily take electrons from other substances.

Species with very negative values are often good reducing agents because their reduced forms readily lose electrons. This comparison helps explain why zinc can react with copper ions, while copper metal does not usually reduce zinc ions.

In a working galvanic cell, electrons travel through the wire from the anode to the cathode. The anode metal may dissolve as atoms become ions, which reduces the mass of that electrode. At the cathode, ions in solution can gain electrons and form a solid coating, which increases electrode mass.

The salt bridge has an essential job. It lets ions move between the half-cells so charge does not build up.

Without this ion movement, one solution would become too positive and the other too negative. Electron flow would quickly stop even if the chemical reaction could still occur.

Students often make errors because they treat voltage like an amount of substance. A balanced equation may need two, three, or more copies of a half-reaction to make electron numbers match. This changes the number of electrons transferred and changes the free energy for the full reaction.

It does not change the listed potential for that half-reaction. Potential measures energy per unit charge, so it stays the same when the reaction is scaled up. Keep two separate ideas in your work.

Balance electrons before adding half-reactions. Then use the original table values once to find the cell voltage. This habit prevents a very common calculation mistake.

Real cells rarely operate under standard conditions. Ion concentrations change as a battery discharges, gases may not be at the reference pressure, and temperature can differ from room temperature. The Nernst equation adjusts the cell voltage for these conditions.

It uses the reaction quotient, which compares the concentrations or pressures of products with those of reactants after the overall equation is balanced. Pure solids and pure liquids are left out because their effective concentration stays nearly constant. As products build up, the forward reaction often becomes less favorable and the voltage falls.

This is one reason batteries weaken during use. A cell at equilibrium has no net driving force, so its voltage is zero. When practicing, write the balanced overall reaction first, identify which substances belong in the reaction quotient, and check that the electron count matches the free energy calculation.