Write the net cell equation for this electrochemical cell. phases are optional. do not include the concentrations. sn(s)||sn2+(aq, 0.0155 m)

Question

Write the net cell equation for this electrochemical cell. phases are optional. do not include the concentrations. sn(s)||sn2+(aq, 0.0155 m)∥∥ag+(aq, 3.50 m)||ag(s) net cell equation: sn +2ag^{+}->sn^{2+} +2ag sn+2ag+⟶sn2++2ag special δσω λμπ reset ( ) [ ] xyxyyyx⟶↽−−⇀ • (s) (l) (aq) (g) calculate e∘cell, δg∘rxn, δgrxn, and ecell at 25.0 ∘c, using standard potentials as needed

in progress 0

Eu00b0cell = 0.94 V

Ecell = 1.00 V

u0394G = -1.9 x 10u2075 J

u0394Gu00b0 = -1.8 x 10u2075 J

Explanation:

Let’s consider this electrochemical cell:

Sn(s)|Snu00b2u207a(aq,0.0155M)||Agu207a(aq, 3.50M)|Ag(s)

The corresponding half-reactions are:

Oxidation (anode): Sn(s) u2192 Snu00b2u207a(aq) + 2 eu207b Eu00b0red = -0.14 V

Reduction (cathode): 2 Agu207a(aq) + 2 eu207b u2192 2 Ag(s) Eu00b0red = 0.80 V

The standard cell potential (Eu00b0cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

Eu00b0cell = Eu00b0red, cat – Eu00b0red, an = 0.80 V – (-0.14 V) = 0.94 V

We can find the cell potential using the Nernst equation.

Ecell = Eu00b0cell – (0.05916/n) . log Q

Ecell = 0.94 V – (0.05916/2) . log ([Snu00b2u207a]/[Agu207a]u00b2)

Ecell = 1.00 V

We can find u0394G and u0394Gu00b0 using the following expressions.

u0394G = -n.F.Ecell = (-2mol).(96468J/mol.V).(1.00V) = -1.9 x 10u2075 J

u0394Gu00b0 = -n.F.Eu00b0cell = (-2mol).(96468J/mol.V).(0.94V) = -1.8 x 10u2075 J

2. [tex] E^{o}cell = E^{o} (Ag+/Ag) – E^{o}( Sn2+/Sn) [/tex]
where, E^{o} (Ag+/Ag) = std. reduction potential of Ag+ = 0.7994 v
and Sn2+/Sn = std. reduction potential of Sn2+ = -0.14 v

Thus, E^{o}cell = 0.7994v – (-0.14v) = 0.9394 v

Now, u0394G^{o} = -nF[tex] E^{0} [/tex],
where, n = number of electrons = 2
F = Faraday’s constant = 96500 C
u2234u0394G^{o}= 2 X 96500 X 0.9394 = -1.18 X[tex] 10^{5} J/mol[/tex]

Now, using Nernst’s Equation we have,
[tex] [tex]E_{cell} = 0.9394 – frac{2.303X298}{2X96500}log frac{0.0115}{ 3.5^{2} } [/tex]
E_{cell} = 0.9765 v

Finally,u0394G = -nFE = -2 X 96500 X 0.9765 = -1.88 X[tex] 10^{5} J/mol[/tex]