Write the electrocheical half-cell reaction and the corresponding Nernst equatio
ID: 479148 • Letter: W
Question
Write the electrocheical half-cell reaction and the corresponding Nernst equation for each of the lines separating the following phases in Figure 2.11(g):
a. Ti and TiO
b. TiO and Ti2O3
c. Ti2+ and Ti2O3
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 2.2 E(v) 2.2 2 1.8 TiO3. 2H20 1.8 1.6 1.6 1.4 E (b) 1.4 1.2 1.2 0.8 0.8 0.6 0.6 0.4 0.4 0.2 TiO2 la) 0.2 0.2 NTI 0.2 0.4 0.6 0.6 0.8 0.8 1.2 1.2 1.4 1.4 Ti203 1.6 1.6 Tio 1.8 1.8 2.2 -2.4 2.4 L -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 pH (g) TitaniumExplanation / Answer
The general form of Nernst equation is : E= E^0 - 0.059/n log [red/ox]
For Ti/TiO couple, Ti is reduced state and TiO is oxidised state.
E = E^0 - 0.059/n log[Ti/TiO]
[Ti (s)] = 1. So the above equation becomes
E = E^0 - 0.059/n log[1/TiO(aq)]
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Ti ==> Ti^2+ + 2e-
2Ti^2+ ==> 2Ti^3+ + 2e-
E = E^0 - 0.059/2log[TiO/Ti2O3]
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E = E^0 - 0.059/2 log[TiO/Ti2O3]
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