Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: will_L on May 10, 2012, 10:31:48 AM
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how to draw the cell diagram of cl2.
Given e.m.f of the cell at standard conditions is +1.36v
reactions at the electrodes are: H(g) = (H+)(aq) + e
and Cl2(g) + (2e-) = (2Cl-)(aq)
giving an overall cell reaction : Cl2(g) + H2(g) = (2Cl-)(aq) + (2H+)(aq)
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I would guess that you would use platinum electrodes in both of the cells.
Pt|H2, H+||Cl2, Cl-|Pt
http://en.wikipedia.org/wiki/Standard_hydrogen_electrode
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yeah, but why is it Cl2(g) beside the salt bridge(//) instead of Cl- ? i thot the ions suppose to be next to the salt bridge.
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yeah, but why is it Cl2(g) beside the salt bridge(//) instead of Cl- ? i thot the ions suppose to be next to the salt bridge.
I wrote the ions by the platinum electrode because that is where reduction occurs, at the cathode. The Cl2 is reduced to Cl- ions at the cathode and H2 is oxidized to H+ at the anode.
I'm not very familiar with line notation for galvanic cells, so my notation might be wrong. I wrote it so that you could "read" the reaction from left to right (H2 is oxidized to H+ and Cl2 is reduced to Cl-).
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it is correct, cl2(g) and cl- shud be seen as a whole. thanks a lot.