5. (5 points) Consider the reaction: Mg (s) + 2 Ag+ (aq) Mg2 + (aq) + 2 Ag (s) E

Question

5. (5 points) Consider the reaction: Mg (s) + 2 Ag+ (aq) Mg2 + (aq) + 2 Ag (s) Eo cell = 3.17% Calculate E cell for the related reaction: 2+ 6. (5 points) Consider an electrochemical cell constructed from the following half-cells, linked by a KCl salt bridge. Note that both half-cells are written as reductions, and you will need to determine the cell reaction and its direction when "spontaneous". . Fe2+ (aq) + 2e. Fe (s) +2n2+ (aq) + 2e- Zn (s) [ (Fe2+]- 1 M (standard conditions) [ (Zn2+]-1 M (standard conditions) When the cell is running spontaneously, write the balanced cell reaction, and calculate E cell at 25 c.

Explanation / Answer

Q5

even though the reaction has twice stoichiometric relationship, the reduction potential remains the SAME

Since E°cell wont depend on the stoichioemtric species, given concentrations remain the same, ie. Q ratio is kept

Q6

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Fe2+ + 2 e Fe(s) 0.44;

Zn2+ + 2 e Zn(s) 0.7618

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = EFe - En = (-0.44) --0.7618 = 0.3218 V

the reaction:

Fe+2 + Zn(s) = Fe(s) + Zn+2(aq)

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