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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 191
D6C.4 �e relationship between the cell potential and the Gibbs energy change of the
cell reaction, ∆rG = −νFEcell only applies under reversible conditions. �is is
achieved by balancing the cell with an equal and opposite externally applied
potential. Under these circumstances, no current �ows.
Solutions to exercises
E6C.1(b) (i) �e reduction half-reactions, together with their standard electrode po-
tentials from the Resource section, are
R: Ag2CrO4(s) + 2e− → 2Ag(s) +CrO2−4 (aq) E−○(R) = +0.45 V
L: Cl2(g) + 2e− → 2Cl−(aq) E−○(L) = +1.36 V
�e cell reaction is obtained by subtracting the le�-hand reduction half-
reaction from the right-hand reduction half reaction
Ag2CrO4(s) + 2Cl−(aq)→ 2Ag(s) +Cl2(g) +CrO2−4 (aq)
�e standard cell potential is calculated as the di�erence of the two stan-
dard electrode potentials [6D.3–224], E−○cell = E−○(R) − E−○(L)
E−○cell = (+0.45 V) − (+1.36 V) = −0.91 V
(ii) �e reduction half-reactions and their standard electrode potentials are
R: Sn4+(aq) + 2e− → Sn2+(aq) E−○(R) = +0.15 V
L: Fe3+(aq) + e− → Fe2+(aq) E−○(L) = +0.77 V
�e cell reaction is obtained by subtracting the le�-hand reduction half-
reaction from the right-hand reduction half-reaction, a�er �rst multiply-
ing the le�- hand half-reaction by two so that the numbers of electrons in
both half-reactions are the same.
Sn4+(aq) + 2Fe2+(aq)→ Sn2+(aq) + 2Fe3+(aq)
�e standard cell potential is
E−○cell = (+0.15 V) − (+0.77 V) = −0.26 V
(iii) �e reduction half-reactions and their standard electrode potentials are
R: MnO2(s) + 4H+(aq) + 2e− →Mn2+(aq) + 2H2O(l)
L: Cu2+(aq) + 2e− → Cu(s)
with E−○(R) = +1.23 V and E−○(L) = +0.34 V.�e cell reaction (R − L) is
therefore
MnO2(s) + 4H+(aq) +Cu(s)→Mn2+(aq) + 2H2O(l) +Cu2+(aq)
and the standard cell potential is
E−○cell = (+1.23 V) − (+0.34 V) = +0.89 V