Question

Calculate the emf and ΔG for the cell reaction at 298 K
Mg(s)|Mg²⁺(0.1 M) || CU²⁺(0.01 M)|Cu(s)
Given E°cell = 2.71 V
1F = 96,500 C

Answer 1

\[\ce{\underset{\text{(Anode)}}{Mg{(s)}}|Mg^2 +_{(0.1 M)}|| Cu^{2+} _{(0.01 M)}|\underset{\text{cathode}}{Cu_{(s)}}}\]

`"E"_"cell"^circ = 2.71 "V"`

T = 298 K

The cell reaction is :

`"Mg"_((s)) -> "Mg"_("aq")^(2+) + 2"e"`

`("Cu"_("aq")^(2+) + 2"e" -> "Cu"_(("s")))/("Mg"_((s)) + "Cu"_("aq")^(2+) -> "Mg"_(aq)^(2+) + "Cu"_((s)))`

n = 2

Nearest equation is 

`"E"_"cell" = "E"_"cell"^circ + 0.0591/"n"  "log"   ["Cu"_("aq")^(2+)]/"Mg"_(aq)^(2+)`

`= 2.17 + 0.0591/2  "log" 0.01/0.1`

`= 2.17 + 0.0591/2  "log" 10^-1`

`= 2.17 - 0.0591/2  "log" 10`

`= 2.17 - 0.0591/2  xx 1`

`= 2.17 - 0.0591/2 xx 1`

= 2.17 - 0.0296

= 2.140 V

Δ G = - nF E_cell 

= -2 × 96500 × 2.14

= -413020 J

= -413.02 kJ