Question

What is the thermodynamic equilibrium constant under standard conditions for the following balanced redox reaction? Zr(s) + O2(g) - ZrO2 (s) Een=2.463 V

Answer 1

Equilibrium constant =
`2.23 * 10^(83)`

Step-by-step explanation:

`Zr(s) + O_2(g) \rightarrow ZrO_2(s)`

`E^0_(cell)` = 2.463 V

Equilibrium constant is related with
`E^0_(cell)` as

`E_(cell)=E^0_(cell) - (2.303 RT)/(nF) ln k_(eq)`

In standard condition,

T = 25 °C = 25 + 273 = 298 K

F = 96500 C mol^-1

R = 8.314
`J\ K^(-1)mol^(-1)`

On substituting values, the above expression becomes:

`E_(cell)=E^0_(cell) - (0.059)/(n) log k_(eq)`

n = 2

At equilibrium,
`E_(cell)= 0`

`0=E^0_(cell) - (0.059)/(2) log k_(eq)`

`log k_(eq)=(2 * 2.463)/(0.059)`

= 83.35

`K_(eq) = antilog 83.35 = 2.23 * 10^(83)`