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A galvanic (voltaic) cell consists of an electrode composed of aluminum in a 1.0 M aluminum ion solution and another electrode...Asked by Anonymous
A galvanic (voltaic) cell consists of an electrode composed of chromium in a 1.0 M chromium(II) ion solution and another electrode composed of gold in a 1.0 M gold(III) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C.
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Answered by
DrBob222
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.
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.
Answered by
Devron
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.
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.
Answered by
Devron
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
E(cell)=E(cathode)-E(anode)= 1.42-(-0.41)= 2.83V
Answered by
Devron
1.83V not 2.83V
Answered by
anonymous
answer: 2.24V
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