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
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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.

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Accepted Answer

### First Question (Half - reaction identification)
# Explanation:
## 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.
# Answer: oxidation

### Second Question (Identifying oxidized element)
# Explanation:
## 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: Mg

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QUESTION IMAGE

Question

Explanation:

First Question (Half - reaction identification)

Step1: Recall oxidation/reduction definitions

Step2: Analyze electron change

Step1: Recall oxidation definition

Step2: Determine oxidized element

Answer:

Second Question (Identifying oxidized element)