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 Reactions that might form oh- ion.

When a pink color develops around a metal in a gel containing phenolphthalein indicator, it means that the solution next to the metal is basic. In an aqueous gel, the pink color means that some hydroxide ions have been formed.

 Although any electrons given up when a reactive metal is oxidized might react at the spot where the oxidation occurs, they can also readily travel to any other spot on the surface of the two joined pieces of metal. That means, it is possible that the point where metals atoms are oxidized could be some distance from the point where hydroxide ions are produced.

 Now let's think about what might be most likely to accept these available electrons. Metal atoms typically don't accept electrons to form negatively charged metal ions. Rather, metal ions tend to give up electrons to form positive ions. Things that are easy to reduce have the most positive standard reduction potentials, like halogens, but we don't have any halogens in our system. The gel surrounding the metal consists mainly of water with about one percent of agar. Although water is not easy to reduce, because water has a negative standard reduction potential in basic solution, this substance can be reduced when the reaction is coupled to the oxidation of Zn metal in basic solution, as shown by the following standard reduction potentials:

  Zn ( OH ) 4 2 + 2 e Zn ( s ) + 4 OH size 12{ ital "Zn" \( ital "OH" \) rSub { size 8{4} } rSup { size 8{2 - {}} } +2e rSup { size 8{ - {}} } rightarrow ital "Zn" \( s \) +4 ital "OH" rSup { size 8{ - {}} } } {}

( E ° in 1 MOH = 1 . 28 volts ) size 12{ \( E°ital "in"1 ital "MOH" rSup { size 8{ - {}} } = - 1 "." "28" ital "volts" \) } {}

  2H 2 O + 2 e H 2 ( g ) + 2 OH size 12{2H rSub { size 8{2} } O+2e rSup { size 8{ - {}} } rightarrow H rSub { size 8{2} } \( g \) +2 ital "OH" rSup { size 8{ - {}} } } {}

( E ° in 1 MOH = 0 . 80 volts ) size 12{ \( E°ital "in"1 ital "MOH" rSup { size 8{ - {}} } = - 0 "." "80" ital "volts" \) } {}

 Agar is a polysaccharide (like starch), and polysaccharides are not easy to reduce. Finally we must not forget that the Petri dishes are open to the air, so the agar gel also contains dissolved oxygen, a good acceptor of electrons. At least two reactions involving oxygen deserve serious consideration:

  O 2 ( g ) + H 2 O + 2 e H 2 + OH size 12{O rSub { size 8{2} } \( g \) +H rSub { size 8{2} } O+2e rSup { size 8{ - {}} } rightarrow H rSub { size 8{2} } rSup { size 8{ - {}} } + ital "OH" rSup { size 8{ - {}} } } {}

( E ° in 1 MOH = 0 . 065 volts ) size 12{ \( E°ital "in"1 ital "MOH" rSup { size 8{ - {}} } = - 0 "." "065" ital "volts" \) } {}

  O 2 ( g ) + 2H 2 O + 4 e 4 OH size 12{O rSub { size 8{2} } \( g \) +2H rSub { size 8{2} } O+4e rSup { size 8{ - {}} } rightarrow 4 ital "OH" rSup { size 8{ - {}} } } {}

( E ° in 1 MOH = 0 . 40 volts ) size 12{ \( E°ital "in"1 ital "MOH" rSup { size 8{ - {}} } = - 0 "." "40" ital "volts" \) } {}

The E ° size 12{E°} {} for reduction of oxygen in basic solution is considerably more positive than for the reduction of water. So we definitely must consider the possibility that oxygen might be the species that could most easily be reduced, with OH size 12{ ital "OH" rSup { size 8{ - {}} } } {} (and possibly hydrogen peroxide) being the reduction product.

 The reduction of either water or oxygen produces hydroxide ions, but the formation of a pink color with phenolphthalein does not tell us which reaction might be responsible. Thermodynamics (as measured by the standard reduction potentials) favors reduction of oxygen over reduction of water. However, the reduction of oxygen on many metals is known to have a large activation energy, which usually causes the reaction to be slow. Thus, kinetics may favor the reduction of water, particularly because the concentration of water is much greater than the concentration of oxygen in the agar gel. Can you think of an experiment that might allow you to distinguish if water or oxygen is the major species being reduced? 20-25min.

The golden penny experiment.

  • Your TA will set this up for you by putting 8 g of 30 mesh zinc in the bottom of a 400-mL beaker. It is best to weigh-out the zinc in a watch glass and pour the zinc into a tilted 400-mL beaker so as to keep the zinc on one side of the beaker. Use a spatula and tilt and tap the beaker on the bench top in order to get all the granular zinc to cover about half the bottom of the beaker (one-half not covered: see Figure 5). Carefully pour 200 mL of 1 NaOH down the side of the beaker, being careful not to disturb the distribution of zinc. Use a stirring rod or spatula to clear any remaining granules so that half of the beaker bottom is completely free of zinc granules. Place the beaker on a hot plate in the fume hood and turn the hot plate to medium heat. The solution should be heated to about 80-90 ° C size 12{°C} {} ; if it is heated to boiling the distribution of zinc granules will be disturbed. Continually monitor and check the temperature to keep it in this range. 20-25min.
  • While waiting for the solution to heat, buff six copper pennies with steel wool until they are shiny. Wash them with deionized water and dry. Solder 10-cm lengths of 20-gauge copper wire to two of the pennies, overlapping the wire and penny about 2 to 3 mm from the edge. Solder the free end of one of the copper wires to a 5 x 100 mm strip of zinc metal, as shown in Figure 5. Clean any rosin off the soldered joints with steel wool, and rinse with water.

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Source:  OpenStax, General chemistry lab spring. OpenStax CNX. Apr 03, 2009 Download for free at http://cnx.org/content/col10506/1.56
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