Question

Calculate emf of the following cell at 25°C:

\[\ce{Sn/Sn^2+ (0.001 M) || H+ (0.01 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]

Given: \[\ce{E^\circ(Sn^2+/sn) = -0.14 V, E^\circ H+/H2 = 0.00 V (log 10 = 1)}\]

Answer 1

According to Nernst equation,

`E_"cell"=E_(cell)^0-0.0592/nlog"" ([P])/([R])`

`E_(cell)^0=E_("Oxi(anode)")^0-E_("Oxi(cathode)")^0`

= − 0.14 − 0.0

= −0.14

Reactions:

Anode (oxidation): \[\ce{Sn(s) -> Sn^2+(aq) + 2e^−}\]

Cathode (reduction):  \[\ce{2H+(aq) + 2e− -> H2(g)}\]

Net reaction: \[\ce{Sn(s) + 2H+(aq) -> Sn^2+(aq) + H2(g)}\]

∴ n = 2

\[\ce{[P] = [Sn^2+] = 0.001 M}\]

\[\ce{[R] = [H+] = 0.01 M}\]

So, putting the above values in the formula,

`E_(cell) = -0.14 - 0.0592/2 log([0.001])/([0.01]^2)`

E_cell = −0.1696 V