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Logina level qns electrochem
5 posts
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a current 0.25A is passed thru 400 cm3 of a 5 M solution of NaCl for 35 mins ,using C electrodes. What will be the pH of the solution after the current is turned off?
what is tink is H and Cl ions will be discharged so what is left will be an alkaline solution, but i'm not sure how to go abt calculating the pH... i do not have the answer to this qns.
any help is much appreciated. thanks.
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The ions present are H+, OH-, Na+, Cl-. You have to consider the standard electrode/reduction/oxidation potentials for the reduction and oxidation half equations for the 4 ions.
Between the cations, there is no contest. Na+ is too stable (since Na is too reactive) to be reduced, so 2H+ will be reduced to H2 gas.
Between the anions, you will have to factor in the molarities. Here, the molarity of Cl- is high enough (notice that standard reduction potentials given in the data booklet are for standard conditions, ie. 1 mol/dm3; here we have 5 mol/dm3) such that Cl- will be certainly be preferentially oxidized instead of OH-.
As H+ is reduced and Cl- is oxidized, what is left over will be NaOH sodium hydroxide solution, as arigatoast predicted.
Since NaCl is a neutral salt (product of strong acid and strong alkali), we know the molarity of [H+] to begin with (Note that pH = 7 is neutral only under 25 deg C; the question doesn't state the temperature and pressure, so we assume it's 25 deg C and 1 atm).
Based on the electrons transferred (see secretliker's post above), we can work out exactly how much H+ is removed from solution. Hence we can work out the new no. of mol of H+ and hence the new [H+] of solution after the current is passed, and therefore pH.
Note : Usually for alkaline solutions, we first calculate pOH, and hence obtain pH. In this case, however, we're dealing directly with the reduction half equation that removes protons H+, so no problem here.
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