Redox+equations

A **redox** reaction is a reaction in which both **oxidation** and **reduction** take place. A //half equation// is an equation which shows a substance being oxidised/reduced.

Therefore, in a redox equation we can write two half equations, one for the substance which is oxidised and one for the substance which is reduced. The two half equations for a redox reaction can be combined to give a redox equation.

Eg: the reaction of orange dichromate (VI) ion to the green chromium (III) ion. We start off as follows: (1) Write the the substances involved in the reaction: Cr 2 O 7 2-  (aq) ---> Cr 3+  (aq) (2) Balance the number of atoms of the element that is being oxidised/reduced (in this case it is Cr) We have two Cr on the left but only one on the right so we put a 2 in front of Cr 3+ Cr 2 O 7 2- (aq) ---> 2Cr 3+  (aq) (3) Add water molecules to balance the number of oxygen atoms on both sides. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) ---> 2Cr 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 7H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (l) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">(4) Add hydrogen ions to balance the number of hydrogen atoms on both sides: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 14H+ (aq) ---> 2Cr3+ (aq) + 7H2O (l) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">(5) Add electrons so that the electrical charges on both sides balance: <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The overall charge on the left is (-2 + 14x(+1)) which is +12 while the overall charge on the right side is (2x(+3)) which is +6, thus 6 electrons should be added to the left side so that the charges on both sides balance (+12 -6=+6)
 * How to write half equations? **

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 14H + <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6e - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> ---> 2Cr 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 7H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (l) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">As the electrons appear on the left, it means that the equation is reduction reaction.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Supposing you were given the half equations for a redox reaction. You can join these two half equations to get the full redox reaction equation. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Eg. We have the two half equations for a redox reaction are given below.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 14H + <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6e - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> ---> 2Cr 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 7H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (l) <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Fe 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) ---> Fe 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + e -

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We need to multiply either or both of the equations by a suitable number so that when they are added, the electrons on the two sides cancel each other out. Here, we need to multiply the second equation by 6 so we have 6e - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (the same number of electrons as the first equation)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">6Fe 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) ---> 6Fe 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6e -

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We add the two equations:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 14H + <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6e - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> + 6Fe 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) ---> 2Cr 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 7H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (l) + 6Fe 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6e -

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">If there are H + <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> ions or H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O molecules on both sides, they cancel each other out, so do the e - <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> on both sides of the equation. So you get:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 14H + <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 6Fe 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) ---> 2Cr 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq) + 7H 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O (l) + 6Fe 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> (aq)

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">This is the redox equation. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">In the above equation Cr 2 <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">O 7 2- <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is an oxidising agent (oxidant) as it oxidises Fe 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> to Fe 3+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">In the above equation Fe 2+ <span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is a reducing agent (reductant) as it reduces the dichromate (VI) ion to chromium (III) ion. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">An **// oxidising agent //**<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> is the one that oxidises another substance but is // itself reduced //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">A //** reducing agent **//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> reduces another substance but is itself // oxidised //<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">. <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Whether a substance behaves as an oxidising agent or a reducing agent depends greatly on what it is reacting with.

By the end of this lesson you should be able to:
 * Deduce half equations, given the species involved
 * Deduce the redox equation from the half equation
 * Define the terms oxidising and reducing agent
 * Identify the oxidising and reducing agents in a redox equation

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