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

Answer 1

$E^{o}cell = E^{o} (Ag+/Ag) - E^{o}( Sn2+/Sn)$
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, ΔG^{o} = -nF$E^{0}$,
where, n = number of electrons = 2
F = Faraday's constant = 96500 C
∴ΔG^{o} = 2 X 96500 X 0.9394 = -1.18 X $10^{5} J/mol$

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

Finally, ΔG = -nFE = -2 X 96500 X 0.9765 = -1.88 X $10^{5} J/mol$

Answer 2

Answer:

E°cell = 0.94 V

Ecell = 1.00 V

ΔG = -1.9 × 10⁵ J

ΔG° = -1.8 × 10⁵ J

Explanation:

Let's consider this electrochemical cell:

Sn(s)|Sn²⁺(aq,0.0155M)||Ag⁺(aq, 3.50M)|Ag(s)

The corresponding half-reactions are:

Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻              E°red = -0.14 V

Reduction (cathode): 2 Ag⁺(aq) + 2 e⁻ → 2 Ag(s)    E°red = 0.80 V

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

E°cell = E°red, cat - E°red, an = 0.80 V - (-0.14 V) = 0.94 V

We can find the cell potential using the Nernst equation.

Ecell = E°cell - (0.05916/n) . log Q

Ecell = 0.94 V - (0.05916/2) . log ([Sn²⁺]/[Ag⁺]²)

Ecell = 1.00 V

We can find ΔG and ΔG° using the following expressions.

ΔG = -n.F.Ecell = (-2mol).(96468J/mol.V).(1.00V) = -1.9 × 10⁵ J

ΔG° = -n.F.E°cell = (-2mol).(96468J/mol.V).(0.94V) = -1.8 × 10⁵ J