3. At 298 K, E cell for the Daniell cell is 1.10 V. Determine the equilibrium ratio [Cu²+]/[Zn²+]. [Ans. 6.90 × 10 38]

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The standard cell potential (at 298 K) for the Daniell cell
values of AG and
is 1.10 V. Calculate the corresponding
K and comment on the thermodynamic viability of the
cell reaction:
Zn(s) + Cu²+ (aq) -Zn²+ (aq) + Cu(s)
(F= 96 485 C mol¹; R = 8.314 x 10³ kJ K-¹ mol-¹)
The equation needed is:
AG=-zFE cell
and z is 2 for the cell reaction:
Zn(s) + Cu²+ (aq) → Zn²+ (aq) + Cu(s)
AG = -zFE cell
= -2 x 96485 × 1.10
= -212 267 J per mole of reaction
≈-212 kJ per mole of reaction
In K =
=
AG°
RT
-212
8.314 x 10-3 × 298
In K = 85.6
K = 1.50 × 10³7
The large negative value of AG° and a value of K which is
>>>1 correspond to a thermodynamically favourable reaction,
one which virtually goes to completion.
3. At 298 K, E cell for the Daniell cell is 1.10 V. Determine
the equilibrium ratio [Cu²+]/[Zn²+].
[Ans. 6.90 × 10 387
Transcribed Image Text:The standard cell potential (at 298 K) for the Daniell cell values of AG and is 1.10 V. Calculate the corresponding K and comment on the thermodynamic viability of the cell reaction: Zn(s) + Cu²+ (aq) -Zn²+ (aq) + Cu(s) (F= 96 485 C mol¹; R = 8.314 x 10³ kJ K-¹ mol-¹) The equation needed is: AG=-zFE cell and z is 2 for the cell reaction: Zn(s) + Cu²+ (aq) → Zn²+ (aq) + Cu(s) AG = -zFE cell = -2 x 96485 × 1.10 = -212 267 J per mole of reaction ≈-212 kJ per mole of reaction In K = = AG° RT -212 8.314 x 10-3 × 298 In K = 85.6 K = 1.50 × 10³7 The large negative value of AG° and a value of K which is >>>1 correspond to a thermodynamically favourable reaction, one which virtually goes to completion. 3. At 298 K, E cell for the Daniell cell is 1.10 V. Determine the equilibrium ratio [Cu²+]/[Zn²+]. [Ans. 6.90 × 10 387
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