Question

use the following information to answer the next question. reduction half - reactions am^{4 + }(aq)+e^{-}→am^{3 + }(aq) e^{o}=+4.60 v tl^{3 + }(aq)+2e^{-}→tl^{+}(aq) e^{o}=+1.25 v ac^{3 + }(aq)+3e^{-}→ac(s) e^{o}=-2.20 v cs^{+}(aq)+e^{-}→cs(s) e^{o}=-3.03 v which of the following equations represents a spontaneous oxidation - reduction reaction? a. 2ac^{3 + }(aq)+3tl^{+}(aq)→3tl^{3 + }(aq)+2ac(s) b. tl^{3 + }(aq)+2am^{3 + }(aq)→2am^{4 + }(aq)+tl^{+}(aq) c. cs(s)+am^{4 + }(aq)→am^{3 + }(aq)+cs^{+}(aq) d. ac(s)+3cs^{+}(aq)→3cs(s)+ac^{3 + }(aq)

Explanation:

Step1: Recall spontaneity condition

A spontaneous redox - reaction has a positive standard cell potential ($E_{cell}^0> 0$). The standard cell potential is given by $E_{cell}^0 = E_{cathode}^0−E_{anode}^0$, where the reduction half - reaction occurs at the cathode and the oxidation half - reaction occurs at the anode.

Step2: Analyze each option

Option a

The reduction half - reactions are:
$Ac^{3 + }(aq)+3e^-\to Ac(s)\quad E^0=- 2.20\ V$ and $Tl^{+}(aq)+e^-\to Tl(s)\quad E^0 = + 1.25\ V$.
The overall reaction $2Ac^{3 + }(aq)+3Tl(aq)\to3Tl^{+}(aq)+2Ac(s)$ has $E_{cell}^0=E_{cathode}^0 - E_{anode}^0=1.25-(-2.20)=3.45\ V>0$. This is a spontaneous reaction.

Option b

The half - reactions are: $Tl^{3 + }(aq)+2e^-\to Tl^{+}(aq)\quad E^0 = + 1.25\ V$ and $Am^{3 + }(aq)+e^-\to Am^{4 + }(aq)\quad E^0= + 4.60\ V$.
$E_{cell}^0=E_{cathode}^0 - E_{anode}^0=4.60 - 1.25 = 3.35\ V>0$. But we need to check all options.

Option c

The half - reactions are: $Cs(s)\to Cs^{+}(aq)+e^-\quad E^0 = 3.03\ V$ (oxidation) and $Am^{4 + }(aq)+e^-\to Am^{3 + }(aq)\quad E^0= + 4.60\ V$ (reduction).
$E_{cell}^0=E_{cathode}^0 - E_{anode}^0=4.60-3.03 = 1.57\ V>0$.

Option d

The half - reactions are: $Ac(s)\to Ac^{3 + }(aq)+3e^-\quad E^0 = 2.20\ V$ (oxidation) and $Cs^{+}(aq)+e^-\to Cs(s)\quad E^0=-3.03\ V$ (reduction).
$E_{cell}^0=E_{cathode}^0 - E_{anode}^0=-3.03 - 2.20=-5.23\ V<0$. This is a non - spontaneous reaction.

Answer:

a