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Untitled - Kelly Walsh High School

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270 CHEMISTRY FOR THE UTTERLY CONFUSED<br />

If there is an inert electrode present, then show where the inert electrode is with<br />

its phase boundary. If the electrode components are in the same phase, then<br />

separate them by commas; if not, a vertical phase boundary line. For example,<br />

consider the following redox reaction:<br />

Ag (aq) Fe 2 (aq) l Fe 3 (aq) Ag(s)<br />

The oxidation of the ferrous ion to ferric doesn’t involve a solid, so we must use<br />

an inert electrode, such as platinum. The cell notation would then be:<br />

Pt(s) | Fe 2 (aq), Fe 3 (aq) || Ag (aq) | Ag(s)<br />

18-3 Standard Reduction Potentials (E)<br />

In the discussion of the Daniell cell, we indicated that this cell produces a<br />

voltage of 1.10 V. This voltage is really the difference in potential between<br />

the two half-cells. The cell potential (really the half-cell potentials) is<br />

dependent upon concentration and temperature, but initially we’ll simply<br />

look at the half-cell potentials at the standard state of 298 K (25C) and all<br />

components in their standard states (1M concentration of all solutions, 1<br />

atm pressure for any gases and pure solid electrodes). Half-cell potentials<br />

appear in tables as the reduction potentials, that is, the potentials associated<br />

with the reduction reaction. We define the hydrogen half-reaction<br />

(2H (aq) 2e l H 2(g)) as the standard and has been given a value of<br />

exactly 0.00 V. We measure all the other half-reactions relative to it; some<br />

are positive and some are negative. Find the table of standard reduction<br />

potentials in your textbook.<br />

Quick Tip<br />

Here are some things to be aware of in looking at this table:<br />

• All reactions appear in terms of the reduction reaction relative to the standard<br />

hydrogen electrode.<br />

• The more positive the value of the voltage associated with the half-reaction<br />

(E) the more readily the reaction occurs.<br />

• The strength of the oxidizing agent increases as the value becomes more positive,<br />

and the strength of the reducing agent increases as the value becomes more<br />

negative.

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