O 13 points The quinhydrone electrode was introduced in 1921 as a means of measu

Question

O 13 points The quinhydrone electrode was introduced in 1921 as a means of measuring pH. Pt(s) I 1:1 mole ratio of quinone(aq) and hydroquinone(aq), unknown pH ll cl-dag, o.50 l Hgzcl2(so IHgon I Ptus) The solution whose pH is to be measured is placed in the left half-cell, which also contains a 1:1 mole ratio of quin (ceHeo2) and hydroquinone (ceHeozo. The half-cell reaction is given below. O 2H+ 2e- at HO OH Quinone Hydroquinone (a) write reduction half reactions for each half-cell. (use the lowest possible whole number coefficients. Omit states of matter from your answer) Help Right half-cell: 9 Help Select the Nernst equation for each half-cell. Left half-cell: E 0.268 0.05916 log [Cl

Explanation / Answer

Left half cell

Q+2H++2e------->QH2

Right:

HgCl2+2e-------->2Hg(l)+2Cl-

Nernst equation for left cell

E=0.7-0.05916/2 log([C6H6O2]/([C6H4O2][H+]2))

Right cell

E=0.268-0.05916/2 log([Cl-]2)

Ecell=E(right)-E(left)=0.268-0.05916/2 log([Cl-]2)-(0.7-0.05916/2 log([C6H6O2]/([C6H4O2][H+]2)))

Ecell=-0.432-0.05916/2 log(([C6H4O2][H+]2[Cl-]2/[C6H6O2])

[C6H6O2]=[C6H4O2]

Cl=0.50 M

So Ecell=-0.432+0.0178 -0.05916 log[H+]

we know that pH=-log [H+]

Ecell=-0.414+0.05916pH

A=-0.414 V

B=0.05916 V/pH

(c)

pH=4.49

Ecell=-0.414+0.05916*4.49=-0.148 V

E<0 so electron flows right to left.

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