'O' & 'A' Level Qn.
so..i had this question (will try to describe as clearly as
possible, cant find diagram)
2 beakers, A and B, contains a bar of zinc and iron each along
with indicators phenolphthalein(detects alkaline solution) and
potassium hexacyanoferrate (detects presence of Fe2+: will turn
blue, otherwise colorless). the metals are linked by a wire and a
voltage is observed. a salt bridge connects the 2 beakers together,
with the salt bridge being partially submerged in both beakers'
indicators.
the question was "would any color change be observed at the iron
electrode?EYA"
so, i reasoned out this way
zinc is more reactive than iron.
zinc will reduce iron, so iron will gain electrons.
Fe + 2e- -> Fe2- (wtf?)
the answer is that color change will be detected (meaning there
is Fe2+ present), but im quite lost as to how Fe2+ is actually
obtained......
pls feel free to clear any misconceptions, i'm not very adept
with this topic 
UltimaOnline's comments :
The question is obviously incomplete (thus the answer
is open to possibilities) because the electrolytes in
both compartments (connected by the salt bridge) are not
specified.
If instead, we modify the question to :
2 beakers, A and B, contain a bar of zinc and iron each
along with indicators phenolphthalein (detects alkaline solution)
and potassium hexacyanoferrate (detects presence of Fe2+: will turn
blue, otherwise colorless). The metals are linked by a wire and a
voltage is observed. A salt bridge connects the 2 beakers
together, with the salt bridge being partially submerged in both
beakers' indicators. A colour change was observed at the cathodic
electrolyte. Suggest and explain two possible scenarios (including
identities of electrolytes) that would allow this.
Solution :
One possible scenario, would have the cathodic
electrolyte as an aqueous Fe3+ solution.
Assuming unimolarities (ie. standard conditions), the standard
reduction potential of the cathodic half equation Fe3+ + e-
--> Fe2+ is +0.77V; the standard oxidation potential of the
anodic half equation Zn --> Zn2+ + 2e- is +0.76V; hence a
current would certainly flow and the standard cell potential would
be +1.53V.
3[Fe(CN)6]3- + 4Fe2+
--> Fe4[Fe(CN)6]3
Prussian Red --> Prussian Blue
In such a scenario, the colour change at the cathodic
electrolyte would be due to Prussian Red reacting with
Fe2+(aq) to generate Prussian Blue.
Another possible scenario would have the cathodic
electrolyte as an aqueous acidic solution.
Assuming unimolarities (ie. standard conditions), the standard
reduction potential of the cathodic half equation 2H+ + 2e-
--> H2 is +0.00V; the standard oxidation potential of the
anodic half equation Zn --> Zn2+ + 2e- is +0.76V; hence a
current would certainly flow and the standard cell potential would
be +0.76V.
Notice that H+ ions came from the solvent water, generating OH-
ions as a consequence.
In such a scenario, the colour change at the cathodic
electrolyte would be due to phenolphthalein reacting with
the OH- ions; specifically, di-deprotonation and structural change
from a non-conjugated, tetrahedral center and absorbing UV
frequencies (hence colourless) to a highly conjugated, trigonal
planar center and absorbing blue-green frequencies (hence
pink/fuchsia).
Further reading on the fascinating chemistry of
PrussianRed / PrussianBlue :
http://en.wikipedia.org/wiki/Prussian_Red
http://en.wikipedia.org/wiki/Prussian_blue
Phenolphthalein :
http://en.wikipedia.org/wiki/Phenolphthalein
http://www.science.ca/askascientist/viewquestion.php?qID=534
http://antoine.frostburg.edu/chem/senese/101/features/water2wine.shtml