Half equations

<body topmargin=0 leftmargin=0 marginheight=0 marginwidth=0> <table cellpadding=0 cellspacing=0 border=0 height="100%"><tr> <td valign="top" width=200 BGCOLOR="#003333" TEXT="#F7F7FF" background="03fd4c00dduajq.png"> <div> <TABLE BORDER="0" CELLPADDING="2" CELLSPACING="0" WIDTH="95" BORDERCOLORLIGHT="#C0C0C0" BORDERCOLORDARK="#808080" FRAME="BOX" RULES="ALL" HSPACE="0" VSPACE="0" > <tR> <tD VALIGN=TOP HEIGHT= 46 WIDTH="91"><IMG BORDER="0" SRC="1x1.gif" HEIGHT="46" ALIGN="bottom" WIDTH="91" HSPACE="0" VSPACE="0"></tD> </tR> </TABLE></div> <div> <A HREF="index.htm" TARGET="_top" TITLE="Science For All"><U>home</U></A>  </div> <div> <A HREF="site_map_and_site_search_engine.htm" TARGET="_top" TITLE="Science for all site map"><U>Site map + site search engine</U></A></div> <bR> <div> <A HREF="id44.htm" TARGET="_top" TITLE="KS5 Switchboard"><U><IMG SRC="14a32320.gif" border=0 width="50" height="50" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></U></A> <A HREF="id44.htm" TARGET="_top" TITLE="KS5 Switchboard"><U>KS5 switchboard</U></A></div> <bR> <div> <A HREF="enthalpy.htm" TARGET="_top" TITLE="Enthalpy "><U>Enthalpy </U></A></div> <div> <B><IMG BORDER="0" SRC="Symb075c00b700c8ffffff00.png" HEIGHT="13" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Half equations</B></div> <div> <A HREF="id91.htm" TARGET="_top" TITLE="Autocatalysis"><U>Autocatalysis</U></A></div> <bR> </td> <td valign="top" BGCOLOR="#FFFFFF" TEXT="#080000" bgproperties="fixed" background="05379803.png"> <div> Half Equations</div> <bR> <div> You are visiting the Science For All site</div> <bR> <div> <IMG SRC="0613d050.png" border=0 width="573" height="5" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <bR> <bR> <div> <B>Half equations</B></div> <bR> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Half equations are used to identify and represent <B>redox</B> reactions.</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">The <B>atoms</B> <B>and</B> <B>charges</B> must be <B>balanced</B>.</div> <div> <B><IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Electrons</B> are used to <B>balance</B> the charges on the two sides of the equation.</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">They can be <B>combined</B> to give an overall picture of a reaction or left as individual equations to illustrate the oxidation or the reduction steps separately.</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Half equations are <B>empirical </B>equations. When they are written as a pair, the <B>minimum stoichiometric </B>values should be used. These are then scaled accordingly when combining them.</div> <bR> <div> Consider for example the reaction between magnesium and hydrochloric acid. </div> <bR> <div> The molecular equation is:</div> <bR> <div> Mg + 2HCl <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> MgCl2 + H2</div> <bR> <div> The ionic equation is:</div> <bR> <div> Mg + 2H+ + 2Cl- <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> Mg2+ + 2Cl- + H2</div> <bR> <div> So you see straight away that Cl- is not actually involved since it is present on both sides. The actual reaction would therefore be:</div> <bR> <div> Mg + 2H+ <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> Mg2+ + H2</div> <bR> <div> Now to redox. It is evident that the magnesium has lost electrons since neutral Mg atoms have become Mg2+ cations. </div> <bR> <div> The half equation for this is:</div> <bR> <div> Mg <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> Mg2+ + 2e- </div> <bR> <div> Electrons have been released (Mg oxidised).</div> <bR> <div> The other species that has changed is the H+. The + charge has been removed therefore an electron has been added i.e. reduction. The half equation is:</div> <bR> <div> 2H+ + 2e- <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> H2</div> <bR> <div> When writing the half equations, you deal only with the atoms and ions that are reacting. </div> <bR> <div> The process can also be reversed, you can work back from the half equations and construct the ionic equation. </div> <div> The <B>link</B> is the number of <B>electrons</B>. </div> <bR> <div> See how it works using this example. </div> <div> (i)     Al <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> Al3+ + 3e-</div> <div> (ii)     2H+ + 2e- <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> H2</div> <bR> <div> To combine these equations, the link is the electrons. You would need 3 lots of equation (ii) to 2 lots of equation (i). 6e- would be involved in each case. Thus:</div> <bR> <div> 6H+ + 2Al <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> 2Al3+ + 3H2 </div> <bR> <div> Note that atoms and charges balance. As a check, work out the ionic equation (minus spectator ions) from the (unbalanced) molecular equation:</div> <bR> <div> Al + HCl <FONT SIZE="3" COLOR="#000000" FACE="Wingdings" >à</FONT> AlCl3 + H2</div> <bR> <bR> <div> Practice tasks</div> <bR> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Write half equations for the following:</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">The formation of silver halides from KF, KCl, KBr and KI.</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">The displacement reactions for the following pairs of chemicals Zn/CuSO4; Li/Pb(NO3)2; Fe/AgNO3; Al/FeSO4; Mg/ZnSO4</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">For each of the above examples (displacements), explain, with the aid of <B>suitable equations</B>, why you have ignored certain ions in the half equations.</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Potassium dichromate (K2Cr2O7) acts as an oxidising agent in acidic conditions. In a similar way to the manganate(VII) ion, Cr3+ is produced. Write a balanced half equation to show what is happening.</div> <div> <IMG BORDER="0" SRC="Symb075c00b700f000000000.png" HEIGHT="16" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Sn2+ can be oxidised to Sn4+ in acidic conditions. Write a half equation and use it, and your previous answer, to deduce the number of moles of  potassium dichromate that would oxidise 0.25 moles of a tin(II) salt.</div> </td> </tr></table></body>