What is the equilibrium constant for the
reaction taking place at room temperature
(T = 25◦C) in the battery
Zn(s)| Zn2+(aq)|| Ce4+(aq)| Ce3+(aq) ?
Assume that the number of electrons transferred
in the reaction is n = 2.
Zn2+ + 2 e
− → Zn E
◦
red = −0.76 V
Ce4+ + e
− → Ce3+ E
◦
red = +1.61 V
1. 6.44 × 1077
2. 1.34 × 1080
3. 1.84 × 102
4. 6.52 × 1083
5. 2.37
To find the equilibrium constant (K) for the given reaction, we can use the Nernst equation. The equation relates the equilibrium constant (K) to the standard reduction potentials (E°) of the half-reactions involved in the reaction.
The given reaction can be divided into two half-reactions:
1. Zn2+ + 2e- → Zn (E°red = -0.76 V)
2. Ce4+ + e- → Ce3+ (E°red = +1.61 V)
To use the Nernst equation, we need to calculate the overall cell potential (Ecell). Ecell is the difference between the reduction potentials of the two half-reactions.
Ecell = E°red (Ce4+/Ce3+) - E°red (Zn2+/Zn)
Ecell = (+1.61 V) - (-0.76 V) = +2.37 V (rounded to two decimal places)
The Nernst equation is:
Ecell = E°cell - (RT / nF) * ln(K)
To find the equilibrium constant (K), we rearrange the Nernst equation:
K = exp((E°cell - Ecell) * (nF / RT))
Substituting the values:
E°cell = 2.37 V
Ecell = 2.37 V
n = 2 (number of electrons transferred)
R = 8.314 J/(mol*K) (ideal gas constant)
T = 298 K (25°C converted to Kelvin)
F = 96485 C/mol (Faraday's constant)
K = exp((2.37 - 2.37) * (2 * 96485)/(8.314 * 298))
Simplifying the expression inside the exponential:
K = exp(0)
The exponential of 0 is equal to 1:
K = 1
Therefore, the equilibrium constant for the given reaction is 1.
None of the options (1. 6.44 × 10^77, 2. 1.34 × 10^80, 3. 1.84 × 10^2, 4. 6.52 × 10^83, 5. 2.37) match the calculated equilibrium constant.