Question

question 13 (1 point)
what is the correct balanced redox reaction for
$\ce{al^{3+}(aq) + ag(s) -> al(s) + ag^{+}(aq)}$
$\ce{al^{3+}(aq) + 3ag(s) -> al(s) + 3ag^{+}(aq)}$
$\ce{2al^{3+}(aq) + 3ag(s) -> 2al(s) + 3ag^{+}(aq)}$
$\ce{3al^{3+}(aq) + ag(s) -> 3al(s) + ag^{+}(aq)}$
$\ce{3al^{3+}(aq) + ag(s) -> al(s) + 3ag^{+}(aq)}$
$\ce{al^{3+}(aq) + 3ag(s) -> 3al(s) + ag^{+}(aq)}$

Explanation:

Step1: Analyze oxidation states

Al in \( \text{Al}^{3+} \) has an oxidation state of +3, and in Al(s) it is 0 (reduction, gain of 3 electrons: \( \text{Al}^{3+} + 3e^-
ightarrow \text{Al} \)). Ag in Ag(s) has an oxidation state of 0, and in \( \text{Ag}^+ \) it is +1 (oxidation, loss of 1 electron: \( \text{Ag}
ightarrow \text{Ag}^+ + e^- \)).

Step2: Balance electrons

To balance electrons, the number of electrons lost must equal the number gained. The reduction half - reaction gains 3 electrons, and the oxidation half - reaction loses 1 electron. So we multiply the oxidation half - reaction (Ag) by 3: \( 3\text{Ag}
ightarrow 3\text{Ag}^+ + 3e^- \). The reduction half - reaction remains \( \text{Al}^{3+}+3e^-
ightarrow \text{Al} \).

Step3: Combine half - reactions

Add the two half - reactions together. The electrons (3\( e^- \)) cancel out. So we have \( \text{Al}^{3+}(\text{aq})+3\text{Ag}(\text{s})
ightarrow \text{Al}(\text{s}) + 3\text{Ag}^+(\text{aq}) \). We also check the number of each atom: 1 Al on both sides, 3 Ag on both sides, and the charges: left side charge is +3, right side charge is \( 3\times(+1)= + 3 \), so charge is balanced.

Answer:

\( \boldsymbol{\text{Al}^{3+}(\text{aq}) + 3\text{Ag}(\text{s})
ightarrow \text{Al}(\text{s}) + 3\text{Ag}^+(\text{aq})} \) (the first option: \( \text{Al}^{3+}(\text{aq})+3\text{Ag}(\text{s})
ightarrow \text{Al}(\text{s})+3\text{Ag}^+(\text{aq}) \))