Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
REDOX REACTIONS Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy. Electrochemical processes are redox (oxidation-reduction) reactions in which energy released by a spontaneous reaction is converted to electricity, or in which electricity is used to cause a non- spontaneous reaction to occur. In redox reactions electrons are transferred from one substance to another. The loss of electrons by an element during oxidation is marked by the increase in the element’s oxidation number. In reduction , there is a decrease in oxidation number resulting from a gain of electrons by an element. Balancing Redox Reactions In principle, balancing redox reactions should be no more difficult than balancing any other sort of reaction equation. However, there are some special techniques for handling redox reactions that also give us insight into electron transfer processes. One such procedure is the half-reaction method . In this approach, the overall reaction is divided into two half-reactions, one for oxidation and one for reduction. The equations for the two half- reactions are balanced separately, and then added together to give the overall balanced equation. Given an unbalanced redox reaction equation: 1. Write the unbalanced equation in ionic form 2. Separate the equation into two half reactions 3. Balance each half-reaction for number and type of atoms and charges. For reactions in an acidic medium, add H 2 O to balance the O atoms and H + to balance the H atoms. 4. Add the two half-reactions together and balance the final equation by inspection. The electrons on both sides must cancel. If the oxidation and reduction half-reactions contain different numbers of electrons, we need to multiply one or both half-reactions to equalize the number of electrons. 5. Verify that the equation contains the same type and numbers of atoms and the same charges on both sides of the equation.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
For example, consider the reaction equation showing the oxidation between Fe 2+ ions with dichromate ions Cr 2 O 7 2- in an acidic medium. Stepwise, we have 1. Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2. Oxidation Reduction 3. In the oxidation half-reaction, the atoms are already balanced. To balance the charge, we add an electron to the right hand side of the arrow: Fe 2+ Fe 3+ + e - In the reduction half reaction, since the reaction takes place in an acidic medium, we add seven H 2 O molecules to the right hand side of the arrow to balance the O atoms: Cr 2 O 7 2- 2Cr 3+ + 7H 2 O To balance the H atoms, we add 14 H + ions to the left-hand side: 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O There are now 12 positive charges on the left-hand side, and only six positive charges on the right hand side. Therefore, we add six electrons to the left: 14H + + Cr 2 O 7 2- + 6e - 2Cr 3+ + 7H 2 O 4. In this case we have only one electron for the oxidation half-reaction, and six electrons for the reduction half-reaction, so we need to multiply the oxidation half-reaction by 6:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 20


This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online