Question

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

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