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.
