Question

what chemical reaction is represented by the following half - reaction?
$ag(s) \
ightarrow ag^+(aq) + e^-$
\bigcirc combination
\bigcirc replacement
\bigcirc reduction
\bigcirc oxidation
\bigcirc decomposition
question help: \boxed{video} \boxed{read}
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\bullet question 8
in the reaction below, which element is oxidized?
$mg(s) + pbbr_2(s) \
ightarrow mgbr_2(aq) + pb(s)$
pb
mg
br
none. this isnt a redox reaction.

Explanation:

First Question (Half - reaction identification)

Step1: Recall oxidation/reduction definitions

Oxidation is the loss of electrons, reduction is the gain of electrons. The half - reaction is \(Ag(s)
ightarrow Ag^{+}(aq)+e^{-}\).

Step2: Analyze electron change

In this reaction, silver (\(Ag\)) goes from a neutral state (\(Ag(s)\)) to a \(+ 1\) charged ion (\(Ag^{+}(aq)\)) and loses an electron (\(e^{-}\)). Since oxidation is defined as the loss of electrons, this half - reaction represents an oxidation reaction. Combination reactions involve combining substances, replacement reactions involve one element replacing another, reduction is gain of electrons, and decomposition is breaking down a compound. So this is an oxidation reaction.

Step1: Recall oxidation definition

Oxidation is the loss of electrons (increase in oxidation state). Let's find the oxidation states of each element.

• For \(Mg(s)\): The oxidation state of an element in its elemental form is \(0\).
• For \(Pb\) in \(PbBr_{2}(s)\): Bromine has an oxidation state of \(- 1\) in compounds (except in some cases like peroxides). Let the oxidation state of \(Pb\) be \(x\). In \(PbBr_{2}\), \(x + 2\times(-1)=0\), so \(x = + 2\). In \(Pb(s)\), the oxidation state of \(Pb\) is \(0\).
• For \(Br\) in \(PbBr_{2}(s)\) and \(MgBr_{2}(aq)\): The oxidation state of \(Br\) is \(-1\) in both compounds (no change in oxidation state).
• For \(Mg\) in \(MgBr_{2}(aq)\): Let the oxidation state of \(Mg\) be \(y\). In \(MgBr_{2}\), \(y+2\times(-1) = 0\), so \(y=+2\).

Step2: Determine oxidized element

\(Mg\) goes from an oxidation state of \(0\) (in \(Mg(s)\)) to \(+ 2\) (in \(MgBr_{2}\)). It loses electrons (oxidation state increases), so \(Mg\) is oxidized. \(Pb\) goes from \(+2\) to \(0\) (gain of electrons, reduced), and \(Br\) has no change in oxidation state.

Answer:

oxidation

Second Question (Identifying oxidized element)