Question

A concentration cell based on the following half reaction at 289 K has initial concentrations of 1.39 M , 0.312 M , and a potential of 0.037206 V at these conditions. After 9.1 hours, the new potential of the cell is found to be 0.0095376 V. What is the concentration of at the cathode at this new potential

Answer 1

The concentration at the cathode is
`A_1 = 0.8183M`

Step-by-step explanation:

From the question we are told that

The initial concentration is
`1.39 M \ for \ Zn^(2+) \ at \ anode \ , 0.312 M \ for \ Zn^(2+) \ at \ cathode`

The reaction is

At Cathode
`Zn^(2+) + 2 e^- ------> Zn`

At anode
`Zn -----> Zn ^(2+) + 2e^(-)`

The initial potential is
`+ 0.037206V \ for \ Cathode \ and \ + 0.037206V \ for \ anode`

The complete reaction is

`Zn^(2+) + Zn_((s)) ----> Zn^(2+) + Zn_((s))`

(Cathode) ----------------- (anode)

Looking the reaction above we can see that after the reaction
`Zn^(2+)` at cathode loss some concentration and
`Zn^(2+)` will gain some concentration

Let say the amount of concentration lost/gained is = z M

The after the reaction the concentration of
`Zn^(2+)` at the cathode would be

`A_1 = 1.39 -z`

While the concentration at the anode would be

`A_2 = 0.312 +z`

At initial the condition the net potential of the cell is

`E^i_(net) = 0V`

After time
`t = 9.1 hrs = 9.1 *3600 = 32760s`

The potential of the cell is

`E_(net) = 0.0095376V`

Generally this potential is defined by Nernst equation as

`E_(cell) = E^i_(cell)- (RT)/(nF) ln[([A_2])/([A_2]) ]`

Where R is the gas constant with a value of
`R = 8.314 J/mol \cdot K`

T is the temperature given as
`T = 289K`

n is the number of mole of electron which
`= 2 mol \ e^-`

F is the farad constant with a value of
`= 96500 C /mol\ e^-`

Substituting values

`0.0095376 =0.0 -([8.314][288])/([2] [9600]) ln [((0.312 +z))/((1.39 -z)) ]`

`0.0095376 = 0.12406383 \ ln [(( 0.312 + z))/([1.39 -z]) ]`

`(0.0095376 )/( 0.12406383) = \ ln [(( 0.312 + z))/([1.39 -z]) ]`

`0.0768766 = \ ln [(( 0.312 + z))/([1.39 -z]) ]`

Taking exponent of both sides

`e^(0.0768766 )= [(( 0.312 + z))/([1.39 -z]) ]`

`1.0799 = [(( 0.312 + z))/([1.39 -z]) ]`

`[1.39 -z ]1.0799 = 0.312 + z \\\\1.501 - 1.0799z = 0.312 +z\\\\1.501-0.312 = 1.0799z + z\\\\1.1891 =2.0799z\\\\ z =(1.1891)/(2.0799)\\\\ z =0.5717`

The concentration at the cathode is
`A_1 = 1.39 -0.5717`

`A_1 = 0.8183M`