<html xmlns:v="urn:schemas-microsoft-com:vml" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:w="urn:schemas-microsoft-com:office:word" xmlns:m="http://schemas.microsoft.com/office/2004/12/omml" xmlns="http://www.w3.org/TR/REC-html40">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=us-ascii">
<meta name="Generator" content="Microsoft Word 15 (filtered medium)">
<style><!--
/* Font Definitions */
@font-face
{font-family:"Cambria Math";
panose-1:2 4 5 3 5 4 6 3 2 4;}
@font-face
{font-family:Calibri;
panose-1:2 15 5 2 2 2 4 3 2 4;}
/* Style Definitions */
p.MsoNormal, li.MsoNormal, div.MsoNormal
{margin:0in;
font-size:11.0pt;
font-family:"Calibri",sans-serif;}
span.EmailStyle18
{mso-style-type:personal-compose;
font-family:"Arial",sans-serif;
color:windowtext;}
.MsoChpDefault
{mso-style-type:export-only;
font-size:10.0pt;
font-family:"Calibri",sans-serif;}
@page WordSection1
{size:8.5in 11.0in;
margin:1.0in 1.0in 1.0in 1.0in;}
div.WordSection1
{page:WordSection1;}
--></style><!--[if gte mso 9]><xml>
<o:shapedefaults v:ext="edit" spidmax="1026" />
</xml><![endif]--><!--[if gte mso 9]><xml>
<o:shapelayout v:ext="edit">
<o:idmap v:ext="edit" data="1" />
</o:shapelayout></xml><![endif]-->
</head>
<body lang="EN-US" link="#0563C1" vlink="#954F72" style="word-wrap:break-word">
<div class="WordSection1">
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">There is a mistake in the solutions manual on Carmen (12th, 13th, 14th ed.) for<br>
20.40(b). <o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><br>
20.40(b) The solutions manual uses an electrode not given in the problem. <br>
To get the smallest (+) emf from the 1/2-cells given would <br>
be using the first two half-cells listed, <br>
<br>
Anode: AuBr4^- (aq) + 3 e- ----> Au(s) + 4 Br- (aq) <br>
<br>
Cat: Eu<sup>^3</sup>+ (aq) + 1 e- ----> Eu<sup>^2</sup>+ (aq) <br>
<br>
To balance the electrons you need to multiply the cathode rxn by 3 <br>
<br>
Cat: 3 (Eu^3+ (aq) + 1 e- ----> Eu^2+ (aq)) <br>
<br>
Then you get: <br>
<br>
Anode: AuBr4^- (aq) + 3 e- ----> Au(s) + 4 Br^- (aq) <br>
<br>
Cat: 3 Eu^3+ (aq) + 3 e- ----> 3 Eu^2+ (aq) <br>
<br>
Add them to get the overall rxn. <br>
<br>
AuBr4^- (aq) + Eu^3+ (aq) ----> Au(s) + 4 Br^- (aq) + 3 Eu^2+ (aq) <br>
<br>
E^o = E^o (cathode) - E^o (anode) <br>
<br>
= (-.43 V) - (-.858 V) <br>
<br>
= + 0.43 V <br>
<br>
Dr. Zellmer <o:p></o:p></span></p>
</div>
</body>
</html>