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Chemical Thermodynamics of Tin - Volume 12 - OECD Nuclear ...

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264<br />

A Discussion <strong>of</strong> selected references<br />

ο<br />

Figure A-4: E app,HCl <strong>of</strong> Sn(IV)/Sn(II) vs. c −<br />

(Cl )tot<br />

(experimental data <strong>of</strong> [1914FOR/BAR]<br />

(), solid line, : E = (145.1 ± 1.8) mV).<br />

ο<br />

app,0<br />

(E - (k / 2) log 10<br />

[c Sn(IV)<br />

/ c Sn(II)<br />

] ) / mV<br />

150<br />

140<br />

130<br />

<strong>12</strong>0<br />

110<br />

100<br />

90<br />

80<br />

E° [Sn(IV) / Sn(II)] = (145.1 ± 1.8) mV<br />

b = -(9.4 ± 0.7) mV·M -1<br />

exp. data [1914FOR/BAR]<br />

linear fit vs. c(Cl – ) tot<br />

95% confidence limits<br />

0 1 2 3 4 5 6<br />

c(Cl – ) tot<br />

/ M<br />

Figure A-5: Correction term connecting log 10 {m Sn(IV) /m Sn(II) } and log 10 { m 4 2<br />

Sn<br />

/ mSn<br />

depends on m Cl<br />

− (: actual measurements; : (0,0) schematic calculation using<br />

β = m /{ m 2 m –}<br />

; dashed line: linear extrapolation <strong>of</strong> the data).<br />

+ +<br />

1(2)<br />

SnCl Sn Cl<br />

+ + }<br />

log(1+β 1(2)·m Cl<br />

–) - log(1+β 1(4)·m Cl<br />

–+β 2(4)·m 2 Cl –)<br />

0<br />

-1<br />

-2<br />

-3<br />

-4<br />

-5<br />

schematic calculation using<br />

β 1(2)<br />

= m(SnCl + ) / {m(Sn 2+ )·m(Cl – )}<br />

log β 1(2)<br />

= 1.34 (I = 0, 298.15 K)<br />

β 1(4)<br />

= m(SnCl 3+ ) / {m(Sn 4+ )·m(Cl – )}<br />

log β 1(4)<br />

= 3.59 (I = 5 M HClO4)<br />

β 2(4)<br />

= m(SnCl 2 2+ ) / {m(Sn 4+ )·m 2 (Cl – )}<br />

log β 2(4)<br />

= 6.24 (I = 5 M HClO4)<br />

-6<br />

0 1 2 3 4 5 6<br />

m Cl<br />

– / mol·kg –1<br />

CHEMICAL THERMODYNAMICS OF TIN, ISBN 978-92-64-99206-1, © <strong>OECD</strong> 20<strong>12</strong>

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