Review the SRP (at 25,text{°C}) data in acidi | vedclass

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(C) Given:$Ti^{4+} + e^- 	o Ti^{3+}, \, E^o = -x 	ext{ V}$$Fe^{3+} + e^- 	o Fe^{2+}, \, E^o = -y 	ext{ V}$Since $x  -y$. Therefore,

Vedclass · Accepted Answer

$Ti^{3+} + Fe^{3+} 	o Ti^{4+} + Fe^{2+}$ is a spontaneous process.

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(C) Given:$Ti^{4+} + e^- 	o Ti^{3+}, \, E^o = -x 	ext{ V}$$Fe^{3+} + e^- 	o Fe^{2+}, \, E^o = -y 	ext{ V}$Since $x  -y$. Therefore,$E^o(Ti^{4+}/Ti^{3+}) > E^o(Fe^{3+}/Fe^{2+})$.$1$. $Ti^{4+}$ has a higher reduction potential than $Fe^{3+}$,so $Ti^{4+}$ is a stronger oxidant than $Fe^{3+}$.$2$. $Ti^{3+}$ has a lower reduction potential than $Fe^{2+}$,so $Ti^{3+}$ is a stronger reductant than $Fe^{2+}$.$3$. For the reaction $Ti^{3+} + Fe^{3+} 	o Ti^{4+} + Fe^{2+}$,the cell potential is $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o(Fe^{3+}/Fe^{2+}) - E^o(Ti^{4+}/Ti^{3+}) = -y - (-x) = x - y$. Since $x Thus,the statement in option $C$ is wrong.

Review the $SRP$ (at $25\,\text{°C}$) data in acidic medium:
$Ti^{4+} + e^- \to Ti^{3+}, \, E^o = -x \text{ V}$
$Fe^{3+} + e^- \to Fe^{2+}, \, E^o = -y \text{ V}$
where $x < y$,point out the wrong statement.

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Review the $SRP$ (at $25\,\text{°C}$) data in acidic medium:$Ti^{4+} + e^- \to Ti^{3+}, \, E^o = -x \text{ V}$$Fe^{3+} + e^- \to Fe^{2+}, \, E^o = -y \text{ V}$where $x < y$,point out the wrong statement.