[MrBananaman]

In an upcoming lab we're measuring how pH effects cell potential E. An equal concentration of quinone and hydroquinone is added to a beaker and its cell potential is measured against an Ag/AgCl cell. Easy enough so far.

The hydroquinone is then titrated with Fe(CN)₆³⁺. We do this to find the E of the hydroquinone at whatever the current pH it is, which should be the measured E at 1/2 volume to equivalence point. What I don't understand is why this reaction occurs.

Here are the half reactions:
Fe(CN)₆³⁺ + e^- --> Fe(CN)₆^4-      E^o=.356
Qu + 2H⁺ = 2e^- --> H₂Qu          E^o=.700

Because we're adding the Fe(CN)₆³⁺, it must be the one reduced. However, the E_cell is then:
E=.365-.7 = -.344

And this should mean a non-spontaneous reaction by
ΔG=-nFE

But clearly it's not, or we wouldn't be using it. What am I doing wrong?

[Babcock_Hall]

Think about your first sentence.

[MrBananaman]

Aaah of course it's not spontaneous under standard conditions, but above a certain pH it will be. You can probably even find at what pH it changes using the Nernst equation.

Thank you for the tip!

[Babcock_Hall]

The quinone system is an important one in the mitochondrial electron transport chain, and it has a value of E°' that is near zero.  The prime indicates that this is the biochemical standard state, which is pH 7, but everything else at 1 M.