which of the following equations represents a spontaneous oxidation - reduction reaction? a. 2ac³⁺(aq) +…

which of the following equations represents a spontaneous oxidation - reduction reaction? a. 2ac³⁺(aq) + 3tl⁺(aq) → 3tl³⁺(aq) + 2ac(s) e° = +4.60 v b. tl³⁺(aq) + 2am³⁺(aq) → 2am⁴⁺(aq) + tl⁺(aq) e° = +1.25 v c. cs(s) + am⁴⁺(aq) → am³⁺(aq) + cs⁺(aq) e° = -2.20 v d. ac(s) + 3cs⁺(aq) → 3cs(s) + ac³⁺(aq) e° = -3.03 v use the following information to answer the next question. reduction half - reactions: cs⁺(aq) + e⁻ → cs(s) e° = -3.03 v ac³⁺(aq) + 3e⁻ → ac(s) e° = -2.20 v tl³⁺(aq) + 2e⁻ → tl⁺(aq) e° = +1.25 v am⁴⁺(aq) + e⁻ → am³⁺(aq) e° = +4.60 v remember: soa is the one closest to the top left hand side of table sra is the one closest to the bottom right hand side of table

which of the following equations represents a spontaneous oxidation - reduction reaction? a. 2ac³⁺(aq) + 3tl⁺(aq) → 3tl³⁺(aq) + 2ac(s) e° = +4.60 v b. tl³⁺(aq) + 2am³⁺(aq) → 2am⁴⁺(aq) + tl⁺(aq) e° = +1.25 v c. cs(s) + am⁴⁺(aq) → am³⁺(aq) + cs⁺(aq) e° = -2.20 v d. ac(s) + 3cs⁺(aq) → 3cs(s) + ac³⁺(aq) e° = -3.03 v use the following information to answer the next question. reduction half - reactions: cs⁺(aq) + e⁻ → cs(s) e° = -3.03 v ac³⁺(aq) + 3e⁻ → ac(s) e° = -2.20 v tl³⁺(aq) + 2e⁻ → tl⁺(aq) e° = +1.25 v am⁴⁺(aq) + e⁻ → am³⁺(aq) e° = +4.60 v remember: soa is the one closest to the top left hand side of table sra is the one closest to the bottom right hand side of table

Answer

Explanation:

Step1: Recall the condition for spontaneous redox reaction

A spontaneous oxidation - reduction reaction has a positive standard cell potential ($E_{cell}^o> 0$). The overall cell reaction is composed of two half - reactions, and $E_{cell}^o=E_{cathode}^o - E_{anode}^o$. Oxidation occurs at the anode and reduction occurs at the cathode.

Step2: Analyze each option

Option a

The half - reactions are: Oxidation: $2Ac(s)\to2Ac^{3 + }(aq)+6e^-$ and Reduction: $3Tl^{3+}(aq)+6e^-\to3Tl(aq)$ $E_{cell}^o=E_{cathode}^o - E_{anode}^o=1.25-(- 2.20)=3.45V>0$

Option b

Oxidation: $2Am^{3+}(aq)\to2Am^{4+}(aq)+2e^-$ and Reduction: $Tl^{3+}(aq)+2e^-\to Tl^+(aq)$ $E_{cell}^o=E_{cathode}^o - E_{anode}^o=1.25 - 4.60=-3.35V<0$

Option c

Oxidation: $Am^{4+}(aq)\to Am^{3+}(aq)+e^-$ and Reduction: $Cs^+(aq)+e^-\to Cs(s)$ $E_{cell}^o=E_{cathode}^o - E_{anode}^o=-3.03 - 4.60=-7.63V<0$

Option d

Oxidation: $3Cs(s)\to3Cs^+(aq)+3e^-$ and Reduction: $Ac^{3+}(aq)+3e^-\to Ac(s)$ $E_{cell}^o=E_{cathode}^o - E_{anode}^o=-2.20-(-3.03)=0.83V>0$ But for option a, $E_{cell}^o = 3.45V$ and for option d, $E_{cell}^o=0.83V$. A larger positive $E_{cell}^o$ indicates a more spontaneous reaction.

Answer:

a. $2Ac^{3+}(aq)+3Tl(aq)\to3Tl^{3+}(aq)+2Ac(s)$