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Principles of Instrumental Analysis, 6th ed. Chapter 22
 
 2
 (1) E = 0.771 – 0.0592 log (0.0111/0.0111) = 0.771 V 
 (2) μ = ½[0.0111 × 22 + 0.0111 × 32 +2 × 0.0111 × 12 + 3 × 0.0111 × 12] 
 = 0.100 
 From Table a2-1, 
 2+Fe
γ = 0.40 and = 0.40 × 0.0111 = 0.00444 2+Fe
a
 3+Fe
γ = 0.18 and = 0.18 × 0.0111 = 0.00200 3+Fe
a
 E = 0.771 – 0.0592 log[0.00444/0.00200] = 0.750 V 
22-4. (a) Sn4+ + 2e– Sn2+ E0 = 0.154 V 
 (1) E = 0.154 – (0.0592/2) log[3.00 × 10–5/6.00 × 10–5] = 0.163 V 
 (2) μ = ½[3.00 × 10–5 × 22 + 6.00 × 10–5 × 42 + 6.00 × 10–5 × 12 + 24 ×10–5 × 12] 
 = 6.9 × 10–4
 From Equation a2-3, 
 2+
2 4
Sn 4
0.509(2) 6.9 10log = 0.05085
1 3.28 0.6 6.9 10
γ
−
−
×
− =
+ × ×
 
 2+Sn
 0.890γ = and 2+
5
Sn
0.890 3.00 10a −= × × = 2.67 × 10–5
 4+
2 4
Sn 4
0.509(4) 6.9 10log = 0.195
1 3.28 1.1 6.9 10
γ
−
−
×
− =
+ × ×
 
 4+Sn
γ = 0.638 and = 0.638 × 6.00 × 104+Sn
a –5 = 3.83 × 10–5
 E = 0.154 – (0.0592/2) log[(2.67 × 10–5)/(3.83 × 10–5)] = 0.159 V 
 (b) (1) As in Solution 22-4a, E = 0.163 V 
 (2) Here the contribution of the tin species to μ is neglible and μ = 0.0800