Question 1[Maximum number: 3]Titanium is a transition element in Period 4. It is commonly found as TiO2\mathrm{TiO}_{2}TiO2 in minerals.Acidified Ti3+(aq)\mathrm{Ti}^{3+}(\mathrm{aq})Ti3+(aq) reacts with oxygen dissolved in water as shown.4Ti3++O2+2H2O→4TiO2++4H+ΔG⊖=−436.1 kJ mol−14 \mathrm{Ti}^{3+}+\mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{TiO}^{2+}+4 \mathrm{H}^{+} \quad \Delta G^{\ominus}=-436.1 \mathrm{~kJ} \mathrm{~mol}^{-1}4Ti3++O2+2H2O→4TiO2++4H+ΔG⊖=−436.1 kJ mol−1The standard reduction potential, E⊖E^{\ominus}E⊖, of O2+4H++4e−⇌2H2O\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}O2+4H++4e−⇌2H2O is +1.23 V .Calculate the standard reduction potential, E⊖E^{\ominus}E⊖, in V , of the TiO2+(aq)/Ti3+(aq)\mathrm{TiO}^{2+}(\mathrm{aq}) / \mathrm{Ti}^{3+}(\mathrm{aq})TiO2+(aq)/Ti3+(aq) half-cell. Show your working.E⊖=E^{\ominus}=E⊖=Mark as masteredShow Answer M1: ΔG=−nE⊖cell F AND n=4 M2: ∴E⊖cell =−436100/−4(96500)=1.13 V ecf M3: E⊖cell =E⊖(O2,4H+∣H2O)−E⊖(TiO2+∣Ti3+)=1.23−E⊖(TiO2+∣Ti3+)∴E⊖(TiO2+∣Ti3+)=(+)0.1(V) ecf \begin{aligned} \text { M1: } \Delta G=-n E \ominus_{\text {cell }} F \text { AND } n=4 \text { M2: } \therefore E \ominus_{\text {cell }}=-436100 /-4(96500)=1.13 \mathrm{~V} \text { ecf } \text { M3: } E \ominus_{\text {cell }}=E \ominus\left(O_{2}, 4 H^{+} \mid \mathrm{H}_{2} \mathrm{O}\right)-E \ominus\left(\mathrm{TiO}^{2+} \mid \mathrm{Ti}^{3+}\right)=1.23-E \ominus\left(\mathrm{TiO}^{2+} \mid \mathrm{Ti}^{3+}\right) \therefore E \ominus\left(\mathrm{TiO}^{2+} \mid \mathrm{Ti}^{3+}\right)=(+) \mathbf{0 . 1}(\mathrm{V}) \text { ecf } \end{aligned} M1: ΔG=−nE⊖cell F AND n=4 M2: ∴E⊖cell =−436100/−4(96500)=1.13 V ecf M3: E⊖cell =E⊖(O2,4H+∣H2O)−E⊖(TiO2+∣Ti3+)=1.23−E⊖(TiO2+∣Ti3+)∴E⊖(TiO2+∣Ti3+)=(+)0.1(V) ecf Add to Test