Potenciais de Eletrodo e Equações de Nernst

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 197
(f) �e reduction half-reactions for the cell are
R: MnO2(s) + 4H+(aq) + 2e− →Mn2+(aq) + 2H2O(l)
L: Fe2+(aq) + 2e− → Fe(s)
�e overall cell reaction is
MnO2(s) + 4H+(aq, L) + Fe(s)→Mn2+(aq) + 2H2O(l) + Fe2+(aq)
where H+(aq, L) has been written to emphasise that it is H+ in the le�-
hand compartment that is involved in the reaction. If acid is added to
both compartments then the acid added to the right-hand compartment
will have no e�ect, because H+(aq, R) does not appear in the overall cell
reaction. However, the acid added to the le�-hand compartment will
increase the tendency of the cell reaction as written to move in the for-
ward direction. �erefore ∆rG for the cell will decrease and so Ecell will
increase .
�is is con�rmed by the Nernst equation. Noting that pure solids and
liquids have aJ = 1, the Nernst equation for this cell is
Ecell = E−○cell −
RT
2F
ln
⎛
⎝
aMn2+aFe2+
a4H+(aq,L)
⎞
⎠
�e addition of acid to the le�-hand compartment will increase aH+(aq,L)
whichwill result in Ecell increasing.�e fact that acid has also been added
to the right-hand compartment has no e�ect on Ecell because aH+(aq,R)
does not appear in the Nernst equation.
6D Electrode potentials
Answer to discussion questions
D6D.2 �is is discussed in Impact 10.
Solutions to exercises
E6D.1(b) (i) �e following electrodes are combined
R: Cu2+(aq) + 2e− → Cu(s) E−○(R) = +0.34 V
L: Sn2+(aq) + 2e− → Sn(s) E−○(L) = −0.14 V
�e cell reaction (R−L) is Cu2+(aq)+Sn(s)→ 2Cu(s)+Sn2+(aq), which
is equivalent to the required reaction, and has ν = 2. �e standard cell
potential is given by [6D.3–224], E−○cell = E−○(R) − E−○(L)
E−○cell = (+0.34 V) − (−0.14 V) = +0.48V
�e relationship between the equilibrium constant and the standard cell
potential is given by [6C.5–221], E−○cell = (RT/νF) lnK. Rearranging gives
lnK = νF
RT
E−○cell =
2 × (96485Cmol−1)
(8.3145 JK−1mol−1) × (298 K)
× (+0.48 V) = 37.3...