Posted by Anonymous on Saturday, January 26, 2013 at 1:18pm.
You should have included the reduction potentials because they vary from text to text.
Look up reduction potential for Cr(II) to Cr.
Look up reduction potential for Au(III) to Au.
Choose the more negative one, reverse the reaction and change the sign. Then add the two E values.
Au3+(aq) + 3e --> Au(s) E=+1.42
Cr3+ (aq) + 3e -->Cr(aq) E= -0.74 V
Since the standard-reduction potential is positive for Au, the reaction should occur at the cathode and the reaction for Cr3 will occur at the anode.
E(cell)=E(cathode)-E(anode)= 1.42-(-0.74)= 2.16V
I haven't done this in a very long time, so I'm not sure if this is entirely correct, and I am not sure about the half-cell reaction for Au, since I do not have any a reference that I own and I picked it off a site of the interent.
Sorry, it's Cr3+ (aq) + e -->Cr2+(aq) E= -0.41 V not Cr3+ (aq) + 3e -->Cr(aq) E= -0.74 V
E(cell)=E(cathode)-E(anode)= 1.42-(-0.41)= 2.83V
1.83V not 2.83V
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