Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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Calculate emf of the following cell at 25°C: Sn | Sn2+ (0.001 M) || H+ (0.01 M) | H2 (g) (1 bar) | Pt (s) - Chemistry

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Calculate emf of the following cell at 25°C:

Sn | Sn2+ (0.001 M) || H+ (0.01 M) | H2 (g) (1 bar) | Pt (s)

`E_(Sn^(2+)"/"Sn)^0=-0.14V " " E_((H^+"/"H_2))^0=0.00V`

 

 

 

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Solution

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): Sn(s) → Sn2+(aq) + 2e

Cathode (reduction):  2H+(aq) + 2e → H2(g)

Net reaction: Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)

∴ n = 2

[P] = [Sn2+] = 0.001 M

[R] = [H+] = 0.01 M

So, putting the above values in the formula,

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

Ecell = –0.1696 V

Concept: Galvanic Cells - Introduction
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