Question

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Fe(s) and Cr3 (aq) to give Cr(s) and Fe2 (aq). Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.). Do NOT include spaces, units, punctuation or anything else silly! Use the reduction potentials for Cr3 (aq) is -0.74 V and for Fe2 (aq) is -0.440 V. [a]

Answer 1

The equilibrium constant is
`K = 3.565*10^(-31)`

Step-by-step explanation:

From the question we are told that

The reduction potential of
`Cr^(3-)` =
`-0.74V`

The reduction potential of
`Fe^(2+)` =
`-0.0440`

In order for us to obtain the equilibrium constant we need to understand the process of the reaction

At the cathode
`Cr^(3-)` is reduced to Cr the reaction is

`2Cr^(3-) + 6e^ - ------> 2Cr`

At the anode
`Fe` is oxidized to
`Fe^(2+)` the reaction is

`3Fe -----> 3Fe^(2+) + 6e^-`

The net voltage of the cell is mathematically given as

`E_(cell) = E_(cathode) - E_(anode)`

substituting value

`E_(cell) = -0.74 - (-0.440)`

`= -0.30V`

Next we need to mathematically evaluate the free energy of the cell as

`\Delta G = -n_e F E_(cell)`

where
`n_e` is the number of moles of electron which is 6

F is the farad constant with a value
`F = 96,500 C/mole`

Substituting values

`\Delta G = - ( 6 * 96500 *-0.3)`

`= 17,3700 J/mol`

The equilibrium constant is mathematically represented as

`K = e ^{(\Delta G)/(-(RT)) }`

Where R is the gas constant with value 8.314 J/mol

T is the temperature with a given value of
`T =298K`

Substituting values

`K = e ^{(173700)/(- (8.314 * 298)) }`

`K = 3.565*10^(-31)`