Question

using the activity series, predict if the following reaction will occur, and the chemical formulas of the products that will form.
ni(s) + sn(clo₄)₂(aq) →
options:

• spontaneous, ni(clo₄)₂ and sn
• spontaneous, niclo₄ and sn
• spontaneous, ni₂clo₄ and sn
• spontaneous, nisn and clo₄
• not spontaneous (no reaction)

activity series of metals:
ni(s) → ni²⁺(aq) + 2e⁻
sn(s) → sn²⁺(aq) + 2e⁻
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question 14
using the activity series, predict if the following reaction will occur, and the chemical formulas of the products that will form.
ca(s) + lic₂h₃o₂(aq) →
options:

• spontaneous, cali₂ and c₂h₃o₂
• spontaneous, ca₂c₂h₃o₂ and li
• spontaneous, ca(c₂h₃o₂)₂ and li
• spontaneous, cac₂h₃o₂ and li
• not spontaneous (no reaction)

activity series of metals:
li(s) → li⁺(aq) + e⁻
ca(s) → ca²⁺(aq) + 2e⁻

Explanation:

First Reaction (Ni(s) + Sn(ClO₄)₂(aq) →)

Step1: Determine Redox Tendency

In the activity series, the half - reactions are \(Ni(s)
ightarrow Ni^{2 + }(aq)+2e^-\) and \(Sn(s)
ightarrow Sn^{2 + }(aq)+2e^-\). A more reactive metal will oxidize (lose electrons) and displace the less reactive metal from its compound. To determine if a reaction occurs, we can look at the relative reactivity. The metal that is more easily oxidized (has a more negative reduction potential or is higher in the activity series) will displace the other. For the reaction \(Ni(s)+Sn(ClO_4)_2(aq)
ightarrow\), we can consider the oxidation of \(Ni\) and reduction of \(Sn^{2+}\). The half - reaction for \(Ni\) oxidation is \(Ni
ightarrow Ni^{2+}+2e^-\) and for \(Sn^{2+}\) reduction is \(Sn^{2+}+2e^-
ightarrow Sn\). We need to check the relative reactivity. From the activity series, \(Ni\) is more reactive than \(Sn\) (we can recall the general activity series where \(Ni\) is above \(Sn\) in terms of reactivity for displacement reactions involving \(2+\) ions). So \(Ni\) will displace \(Sn\) from \(Sn(ClO_4)_2\).

Step2: Determine Product Formulas

The compound \(Sn(ClO_4)_2\) has \(Sn^{2+}\) and \(ClO_4^-\) ions. When \(Ni\) displaces \(Sn\), \(Ni^{2+}\) will combine with \(ClO_4^-\) ions. The formula of the compound formed from \(Ni^{2+}\) and \(ClO_4^-\) is \(Ni(ClO_4)_2\) (since the charge of \(Ni^{2+}\) is \(+2\) and \(ClO_4^-\) is \(- 1\), so we need two \(ClO_4^-\) ions to balance the charge: \(Ni^{2+}+2ClO_4^-
ightarrow Ni(ClO_4)_2\)) and the displaced metal is \(Sn\). So the reaction is spontaneous, and the products are \(Ni(ClO_4)_2\) and \(Sn\).

Second Reaction (Ca(s) + LiC₂H₃O₂(aq) →)

Step1: Determine Redox Tendency

The half - reactions are \(Li(s)
ightarrow Li^{+}(aq)+e^-\) and \(Ca(s)
ightarrow Ca^{2+}(aq)+2e^-\). To determine if a displacement reaction occurs, we compare the reactivity of \(Ca\) and \(Li\). In the activity series, \(Li\) is more reactive than \(Ca\) (lithium is higher up in the activity series than calcium). A metal can only displace another metal from its compound if it is more reactive. Since \(Ca\) is less reactive than \(Li\), \(Ca\) cannot displace \(Li\) from \(LiC_2H_3O_2\). So the reaction is not spontaneous (no reaction).

Answer:

For \(Ni(s)+Sn(ClO_4)_2(aq)
ightarrow\): A. Spontaneous, \(Ni(ClO_4)_2\) and \(Sn\)
For \(Ca(s)+LiC_2H_3O_2(aq)
ightarrow\): E. Not Spontaneous (No reaction)