This question is about titanium compounds and ions.
Use data from the table below to identify the species that can be used to reduce TiO2+\text{TiO}^{2+}TiO2+ ions to Ti3+\text{Ti}^{3+}Ti3+ in acidic aqueous solution and no further. Explain your answer.
| Electrode half-equation | Eθ / VE^\theta \text{ / V}Eθ / V |
|---|---|
| TiO2+(aq)+2H+(aq)+e−→Ti3+(aq)+H2O(l)\text{TiO}^{2+}(\text{aq}) + 2\text{H}^+(\text{aq}) + e^- \to \text{Ti}^{3+}(\text{aq}) + \text{H}_2\text{O}(\text{l})TiO2+(aq)+2H+(aq)+e−→Ti3+(aq)+H2O(l) | +0.10+0.10+0.10 |
| Ti3+(aq)+e−→Ti2+(aq)\text{Ti}^{3+}(\text{aq}) + e^- \to \text{Ti}^{2+}(\text{aq})Ti3+(aq)+e−→Ti2+(aq) | −0.37-0.37−0.37 |
| Cu2+(aq)+2e−→Cu(s)\text{Cu}^{2+}(\text{aq}) + 2e^- \to \text{Cu}(\text{s})Cu2+(aq)+2e−→Cu(s) | +0.34+0.34+0.34 |
| Ni2+(aq)+2e−→Ni(s)\text{Ni}^{2+}(\text{aq}) + 2e^- \to \text{Ni}(\text{s})Ni2+(aq)+2e−→Ni(s) | −0.25-0.25−0.25 |
| Mn2+(aq)+2e−→Mn(s)\text{Mn}^{2+}(\text{aq}) + 2e^- \to \text{Mn}(\text{s})Mn2+(aq)+2e−→Mn(s) | −1.18-1.18−1.18 |
Give the oxidation state of titanium in [TiO(H2O)5]2+[\text{TiO}(\text{H}_2\text{O})_5]^{2+}[TiO(H2O)5]2+.
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