Question

)(2) For a Cu2+|Cu concentration cell to have an Ecell of 0.047 V, if the concentration of Cu2* at the cathode is 0.100 M what is the concentration of Cu2* at the anode? a) 0.0025 M b) 0.025 M c) 0.25 M d) 0.10 M e) 4.0 M

Answer 1

`[Cu]_(anode)=0,0025M`

Step-by-step explanation:

The half-equations happening in the cell are

Anode (oxidation)

`Cu_(s)->Cu^(+2)_(aq) +2e^(-) E^(0) =-0,34V`

Cathode (reduction)

`Cu^(+2)_(aq) +2e^(-)->Cu_(s) E^(0) =0,34V`

Then the overall equation

`Cu_(s)+Cu^(+2)_(aq)->Cu^(+2)_(aq)+Cu_(s)`

So

`E^(0 )_(cell) =E^(0 )_(anode)+E^(0 )_(cathode)=0`

For concentration cells, we can calculate the cell potential
`E_(cell)` using the Nernst Equation:

`E_(cell)=E^(0 )_(cell)-(0,0591)/(n) .log([Cu]_(anode))/([Cu]_(cathode))`

Replacing

`0.047=0-(0,0591)/(2) .log([Cu]_(anode))/(0.100 M)`

Then

`[Cu]_(anode)=0,0025M`