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

Answer:

$[Cu]_{anode}=0,0025M$

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:

$E_{cell}=E^{0 }_{cell}-\frac{0,0591}{n} .log\frac{[Cu]_{anode}}{[Cu]_{cathode}}$

Replacing

$0.047=0-\frac{0,0591}{2} .log\frac{[Cu]_{anode}}{0.100 M}$

Then

$[Cu]_{anode}=0,0025M$