The value of the equilibrium constant, K, at 25 °C for the reaction between the pair : Sn(s) and Pb₂ (aq) to give Pb(s) and Sn₂ (aq) is 2.17.
The half reaction are given as follows :
Sn ----> Sn²⁺ + 2e⁻ reductional potential = 0.14 V
Pb²⁺ + 2e⁻ -----> Pb oxidation potential = - 0.13 V
The E° cell = 0.14 - 0.13
= 0.01 V
using the formula given as :
E° cell = 0.0592 / n (log K)
where n = number of electron
0.01 = 0.0592 / 2 (log K)
k = 2.17
The equilibrium constant K = 2.17
Thus, The value of the equilibrium constant, K, at 25 °C for the reaction between the pair : Sn(s) and Pb₂ (aq) to give Pb(s) and Sn₂ (aq) is 2.17.
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