April 21, 2019, 04:42:42 AM
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### Topic: Redox tritation , potential calculation doubt  (Read 304 times)

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##### Redox tritation , potential calculation doubt
« on: January 30, 2019, 02:57:47 PM »
I don't understand why the Potential E for the tritation of H3AsO3 with Ce4+ at Φ<1 (excess of H3AsO3)is:

E= E°(H3AsO4/H3AsO3) +(0.059/2)log {([H3AsO4][H+]2)/ [H3AsO4]}

At Φ<1 I can assume E(tritrand) >>E(tritrant). Here the tritand is H3AsO3 (H3AsO3---> H3AsO4)

So E= E cat-E anod ≈ -Eanod

So why my textbook says that the potential is:

E= E°(H3AsO4/H3AsO3) +(0.059/2)log {([H3AsO4][H+]2)/ [H3AsO4}

Or in other words E= E(H3AsO4-->H3AsO3)   (reduction potential)
Shouldn't be E = -E(H3AsO4--> H3AsO3),because H3AsO3/H3AsO4 is the anode!!?

??
Thanks!

#### Vidya

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##### Re: Redox tritation , potential calculation doubt
« Reply #1 on: February 04, 2019, 01:08:44 AM »

Or in other words E= E(H3AsO4-->H3AsO3)   (reduction potential)
Shouldn't be E = -E(H3AsO4--> H3AsO3),because H3AsO3/H3AsO4 is the anode!!?

??
Thanks!

yes it is oxidation and if reduction potential is given then you need to take reverse of it.
If the formula is Ecell = Ecathode + E anode
however is you are using this formula
Ecell = Ecathode - E anode
Then you need to take reduction potential as such as negative of the formula is making it oxidation potential.