Question

half-reaction | e° (v)
au³⁺ + 3e⁻ → au | + 1.42
ti²⁺ + 2e⁻ → ti | -1.63
cu²⁺ + e⁻ → cu¹⁺ | + 0.16
sn²⁺ + 2e⁻ → sn | - 0.14
rank these reactions in order from most to least likely to be reduced.
options:
au³⁺ > ti²⁺ > cu²⁺ > sn²⁺
ti²⁺ > sn²⁺ > cu²⁺ > au³⁺
ti²⁺ > au³⁺ > cu²⁺ > sn²⁺
au³⁺ > cu²⁺ > sn²⁺ > ti²⁺

Explanation:

In electrochemistry, the standard reduction potential ($E^\circ$) indicates the tendency of a species to be reduced. A more positive $E^\circ$ value means the species is more likely to be reduced. The $E^\circ$ values are: $\ce{Au^3+}$: +1.42 V, $\ce{Cu^2+}$: +0.16 V, $\ce{Sn^2+}$: -0.14 V, $\ce{Ti^2+}$: -1.63 V. Ranking from most positive (most likely to be reduced) to least positive (least likely to be reduced): $\ce{Au^3+}$ (1.42) > $\ce{Cu^2+}$ (0.16) > $\ce{Sn^2+}$ (-0.14) > $\ce{Ti^2+}$ (-1.63).

Answer:

D. $\ce{Au^3+ > Cu^2+ > Sn^2+ > Ti^2+}$ (assuming the last option is labeled D; if original labels differ, adjust, but the correct order is $\ce{Au^3+ > Cu^2+ > Sn^2+ > Ti^2+}$)