Question

What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cu2+ concentration is 1.44 M and the Mg2+ concentration is 1.47×10-4 M ?

Cu2+(aq) + Mg(s) Cu(s) + Mg2+(aq)

Answer:  V

The cell reaction as written above is spontaneous for the concentrations given:
_______
true
false

Answer 1

Lets find Eo 1st
from data table:
Eo(Mg2+/Mg(s)) = -2.372 V
Eo(Cu2+/Cu(s)) = 0.337 V

As per given reaction/cell notation,
cathode is (Cu2+/Cu(s))
anode is (Mg2+/Mg(s))

Eocell = Eocathode - Eoanode
= (0.337) - (-2.372)
= 2.709 V

Number of electron being transferred in balanced reaction is 2
So, n = 2

use:
E = Eo - (2.303*RT/nF) log {[Mg2+]^1/[Cu2+]^1}

Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591

So, above expression becomes:
E = Eo - (0.0591/n) log {[Mg2+]/[Cu2+]}
E = 2.709 - (0.0591/2) log ((1.47*10^-4)/1.44)
E = 2.709-(-0.118)
E = 2.827 V
Answer: 2.83 V

Since E is positive, this is spontaneous
Answer: True