Question

3) Consider a voltaic cell in which the following reaction occurs: Zn(s) + Sn2+(aq)  Zn2+(aq) + Sn(s)

a) Calculate Eo for the cell.
b) when the cell operates, what happens to the concentration of Zn2+? The concentration

of Sn2+ ?

c) When the cell voltage drops to zero, what is the ratio of the concentration of Zn2+ to that of Sn2+ ?

d) If the concentration of both cations is 1.0 M originally, what are the concentrations when the voltage drops to zero?

Answer 1

3. (a)

Zn(s) + Sn²⁺(aq) Zn²⁺(aq) + Sn(s)

Zn(s) Zn²⁺(aq) + 2e⁻ E⁰ = + 0.76 V ----------------------(1)
Sn²⁺(aq) + 2e⁻ Sn(s) E⁰ = - 0.14 V ----------------------(2)

Adding equation (1) and (2), we get;

Zn(s) + Sn²⁺(aq) Zn²⁺(aq) + Sn(s) ;   E⁰ = + 0.76 V + ( - 0.14 V) = 0.62 V

(b)

In the voltaic cell E⁰ has a positive value and hence, indicating a spontaneous reaction. So, concentration of Zn²⁺ will increase and the concentration of Sn²⁺ will decrease.

(c)

Using Nerst equation;

E_cell = E⁰ - (RT/nF) lnQ

Q = Zn²⁺(aq) / Sn²⁺(aq)

So, E_cell = E⁰ - (RT / nF) ln [Zn²⁺(aq) / Sn²⁺(aq)]

E_cell = 0
R = 8.314 J K^-1 mol^-1
T = 25 ^oC = 298 K
n = 2
F = 96485 C mol^-1

Substituting the values, we get

0 = 0.62 V - [ { (8.314 J K^-1 mol^-1) (298 K) } / {2 x 96485 C mol^-1} ] ln [Zn²⁺(aq) / Sn²⁺(aq)]

0 = 0.62 V - 0.01 V ln [Zn²⁺(aq) / Sn²⁺(aq)]

0.01 V ln [Zn²⁺(aq) / Sn²⁺(aq)] = 0.62 V

ln [Zn²⁺(aq) / Sn²⁺(aq)] = 0.62 / 0.01

ln [Zn²⁺(aq) / Sn²⁺(aq)] = 62

[Zn²⁺(aq) / Sn²⁺(aq)] = e⁶² = 8.44 x 10²⁶

(d)

Initial concentration :
[Zn²⁺(aq) / Sn²⁺(aq)] = 1/1

Concentration when Ecell = 0
[Zn²⁺(aq) / Sn²⁺(aq)] = 8.44 x 10²⁶

[Zn²⁺(aq) / Sn²⁺(aq)] = (8.44 x 10²⁶) / 1