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There is a mistake in the solutions manual on Carmen (12th-14th ed.)
for 20.40(b).<br>
<br>
20.40(b) The solutions use 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 class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+
(aq) + 1 e- ----> Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
(aq) <br>
<br>
To balance the electrons you need to multiply the cathode rxn by
3 <br>
<br>
Cat: 3 (Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+
(aq) + 1 e- ----> Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
(aq)) <br>
<br>
Then you get: <br>
<br>
Anode: AuBr4^- (aq) + 3 e- ----> Au(s) + 4 Br- (aq) <br>
<br>
Cat: 3 Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+
(aq) + 3 e- ----> 3 Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
(aq) <br>
<br>
Add them to get the overall rxn. <br>
<br>
AuBr4^- (aq) + Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+
(aq) ----> Au(s) + 4 Br- (aq) + 3 Eu<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
(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
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