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What is the ratio of the Sn ' concentrations in th

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Answer #1

first consider the oxidation reaction

Sn ---> Sn+2 (ox) + 2e-

reduction reaction :

Sn+2 (red) +2e- ---> Sn

net reaction is


Sn + Sn+2 (red) --> Sn+2 (ox) + Sn


now

the reacton quotient is given by

Q = [Sn+2 (ox) ] / [Sn+2 (red)

now

according to nernst equation

E = Eo - ( 0.0592/n) log Q

here

n = 2 as two electrons are transferred

also

Eo = 0 for concentration cells

given

E = 0.18

so

0.18 = 0 - ( 0.0592/2) log Q

Q = 8.3 x 10-7

now

Q = [Sn+2 (ox) ] / [Sn+2 (red) = 8.3 x 10-7

so

[Sn+2 (ox) ] / [Sn+2 (red) = 8.3 x 10-7

so

the ratio is 8.3 x 10-7

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