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

• Anode: Sn → Sn²⁺ + 2e⁻ (oxidation)

• Cathode: 2H⁺ + 2e⁻ → H₂ (reduction)

\[\ce{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.00V - (-0.14 V) = 0.14 V}\]

`E_(cell) = E_(cell)^\circ - 0.0591/n log Q`

Where:

• n = 2 (electrons exchanged)

• Q = reaction quotient

\[\ce{Sn(s) + 2H+ -> Sn^2+ + H 2 (g)}\]

`Q = ([Sn^(2+)][H2])/[H^+]^2 = (0.001xx1)/(0.01)^2 = 0.001/0.0001 = 10`

`E_(cell) = 0.14 - 0.0591/2 log(10) = 0.14 - 0.0591/2 xx 1 = 0.14 - 0.02955`

= 0.11045 V

E_cell = 0.110 V