Homework Answers
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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