Question

A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
Calculate the cell potential at 25 ∘C under each of the following conditions.

A) Standard condition

B) [Sn2+]= 1.64×10⁻² M ; [Mn2+]= 2.25 M

C) [Sn2+]= 2.25 M ; [Mn2+]= 1.64×10⁻² M

Answer 1

A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
Calculate the cell potential at 25 ∘C under each of the following conditions.

A) Standard condition

B) [Sn2+]= 1.64×10⁻² M ; [Mn2+]= 2.25 M

C) [Sn2+]= 2.25 M ; [Mn2+]= 1.64×10⁻² M

First we calculate the cell potential at 25 ∘C under Standard condition

Reduction potentials for this cell is following:
Sn²⁺(aq) + 2e⁻(aq) → Sn(s) .... E° = -0.14 V
Mn²⁺(aq) + 2e⁻(aq) → Mn(s) .... E° = -1.18 V

Standard cell potential, ΔE°
= [(-0.14) - (1.18)] V
= +1.04 V


B) [Sn2+]= 1.64×10⁻² M ; [Mn2+]= 2.25 M

Here T = 25 ∘C or 298 K

F= 96500 C mol⁻¹.

n = number of electrons, 2
cell potential, ΔE
= ΔE° - (RT / nF) ln ([Mn²⁺] / [Sn²⁺])
= +1.04 - [8.314 × (298) / (2 × 96500)] ln[2.25 / (1.64×10⁻² )] V

= +1.04 -0.0128 ln 137.2

= +1.04 – 0.063
= +0.977 V

C) [Sn2+]= 2.25 M ; [Mn2+]= 1.64×10⁻² M

Here T = 25 ∘C or 298 K

F= 96500 C mol⁻¹.

n = number of electrons, 2
cell potential, ΔE
= ΔE° - (RT / nF) ln ([Mn²⁺] / [Sn²⁺])
= +1.04 - [8.314 × (298) / (2 × 96500)] ln[ (1.64×10⁻² )/ 2.25 ] V

= +1.04 -0.0128 ln 7.289*10^-3

= +1.04 + 0.063
= + 1.10 V