Question

Write the cell reaction and calculate the e.m.f of the following cell at 298 K:

Sn_(s) | Sn²⁺ (0.004 M) || H⁺ (0.020 M) | H_2(g) (1 bar) | Pt_(s)

(Given: `E_(Sn^(2+)"/"Sn)^° = -0.14` V)

Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:

Sn_(s) | Sn²⁺ (0.004 M) || H⁺ (0.02 M) | H_2(g) (1 Bar) | Pt_(s)

(Given: `E_(Sn^(2+)//Sn)^° = -0.14  V, E_(H^(+)//H_2(g), Pt)^° = 0.00  V`)

Answer 1

At cathode: 2H⁺ + 2e → H₂

\[\ce{E^°_{cell} = E^°_{H^{+}/H_2} -  E^°_{Sn^{2+}/Sn}}\]

= 0.00 − (−0.14)

= +0.14 V

Anode:	Sn(s)			→	Sn(aq)2+	+	2e−
Cathode:	2H(aq)+	+	2e−	→	H2(g)
Overall reaction:	Sn(s)	+	2H+	→	Sn(aq)2+	+	H2(g)

`E_"cell" = E_"cell"^° - 0.0591/n log  ([Sn^{2+}])/([H^+]^2)`

`= 0.14 - 0.0591/2 log  ((0.004))/(0.02)^2`

`= 0.14 - 0.0295 log  ((0.004))/((0.0004))`

= 0.14 − 0.0295 log 10

= 0.14 − 0.0295 × 1    ...(∴ log 10 = 1)

= 0.1105 V