Cliffs AP Chemistry, 3rd Edition

Balance charges with electrons.

red: MnO 4 - ++ 85 H+ e-"Mn^2 ++ 4 H O 2

Step 3:Equalize the number of electrons lost and gained. There were 5 e–gained in the reduc-
tion half-reaction, so there must be 5 e–lost in the oxidation half-reaction

ox: 5 Fe2+→5 Fe3++ 5 e–

Step 4:Add the two half-reactions (cancel the electrons).

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ox Fe Fe e

red MnO H e Mn H O

Fe MnO H Fe Mn H O

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85 4

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4. Given the following notation for an electrochemical cell:

Pt(s) | H 2 (g) | H+(aq) || Ag+(aq) | Ag(s)

Which of the following represents the overall balanced (net) cell reaction?

A. H 2 (g) + Ag+(aq) →2 H+(aq) + Ag(s)

B. H 2 (g) + Ag(s) →H+(aq) + Ag+(aq)

C. Ag(s) + H+(aq) →Ag+(aq) + H 2 (g)

D. 2 H+(aq) + Ag(s) →H 2 (g) + Ag+(aq)

E. none of the above

Answer: E

The vertical lines represent phase boundaries. By convention, the anode is written first, at the
left of the double vertical lines, followed by the other components of the cell as they would ap-
pear in order from the anode to the cathode. The platinum is present to represent the presence
of an inert anode. The two half-reactions that occur are

anode: H 2 (g) →2 H+(aq) + 2 e–oxidation
OIL (Oxidation Is Losing electrons)
AN OX (ANode is where OXidation occurs)

cathode: Ag+(aq) + e–→Ag(s) reduction
RIG (Reduction Is Gaining electrons)
RED CAT (REDuction occurs at the CAThode)

In adding the two half-reactions, multiply the reduction half-reaction by 2 so the electrons are
in balance, giving the overall reaction

H 2 (g) + 2 Ag+(aq) →2 H+(aq) + 2 Ag(s)

Reduction and Oxidation