QuestionEnter electrons as $\mathrm{e}^{-}$ . Use smallest possible integer coefficients. If a box is not needed, leave it blank.
For the following electron-transfer reaction: $\mathrm{Ni}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})$
The oxidation half-reaction is: $\square$ $+$ $\square$ $\longrightarrow$ $\square$ $+$ $\square$
The reduction half-reaction is: $\square$ $+$ $\square$ $\longrightarrow$ $\square$ $+$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to identify and write out the oxidation and reduction half-reactions for the given electron-transfer reaction. Watch out! Make sure you correctly identify which element is being oxidized and which is being reduced, and balance the charges in each half-reaction using electrons.

STEP 2

1. Assign oxidation states
2. Identify oxidation and reduction
3. Write half-reactions
4. Balance charges with electrons

STEP 3

Let's assign oxidation states to each element in the reaction.
Remember, pure elements have an oxidation state of zero.
So, $\mathrm{Ni(s)}$ has an oxidation state of zero, and $\mathrm{Cl_2(g)}$ also has an oxidation state of zero.

STEP 4

In the products, $\mathrm{Ni^{2+}(aq)}$ has an oxidation state of +2 (it's given right there!), and each $\mathrm{Cl^{-}(aq)}$ has an oxidation state of -1.

STEP 5

Now, let's see what's happening to our elements.
Nickel ( $\mathrm{Ni}$ ) goes from an oxidation state of zero to +2.
This increase in oxidation state means $\mathrm{Ni}$ is being oxidized!

STEP 6

Chlorine ( $\mathrm{Cl}$ ) goes from an oxidation state of zero to -1.
This decrease in oxidation state means $\mathrm{Cl}$ is being reduced!

STEP 7

The oxidation half-reaction shows what happens to the substance being oxidized.
In our case, it's nickel.
So, we write: $\mathrm{Ni(s)} \longrightarrow \mathrm{Ni^{2+}(aq)}$

STEP 8

The reduction half-reaction shows what happens to the substance being reduced.
Here, it's chlorine.
So, we write: $\mathrm{Cl_2(g)} \longrightarrow 2\mathrm{Cl^{-}(aq)}$ Notice how we added a 2 in front of $\mathrm{Cl^{-}}$ to balance the number of chlorine atoms on both sides.

STEP 9

Now, we need to balance the charges in each half-reaction using electrons ( $\mathrm{e^{-}}$ ).
In the oxidation half-reaction, the left side has a charge of zero, and the right side has a charge of +2.
To balance this, we add two electrons to the right side: $\mathrm{Ni(s)} \longrightarrow \mathrm{Ni^{2+}(aq)} + 2\mathrm{e^{-}}$

STEP 10

In the reduction half-reaction, the left side has a charge of zero, and the right side has a charge of -2 (two chloride ions, each with a -1 charge).
To balance this, we add two electrons to the left side: $\mathrm{Cl_2(g)} + 2\mathrm{e^{-}} \longrightarrow 2\mathrm{Cl^{-}(aq)}$

STEP 11

The oxidation half-reaction is: $\mathrm{Ni}(\mathrm{~s}) \longrightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$ The reduction half-reaction is: $\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})$

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Studdy Solution

STEP 1

STEP 2

1. Assign oxidation states
2. Identify oxidation and reduction
3. Write half-reactions
4. Balance charges with electrons

STEP 3

Let's assign oxidation states to each element in the reaction.
Remember, pure elements have an oxidation state of zero.
So, $\mathrm{Ni(s)}$ has an oxidation state of zero, and $\mathrm{Cl_2(g)}$ also has an oxidation state of zero.

STEP 4

In the products, $\mathrm{Ni^{2+}(aq)}$ has an oxidation state of +2 (it's given right there!), and each $\mathrm{Cl^{-}(aq)}$ has an oxidation state of -1.

STEP 5

Now, let's see what's happening to our elements.
Nickel ( $\mathrm{Ni}$ ) goes from an oxidation state of zero to +2.
This increase in oxidation state means $\mathrm{Ni}$ is being oxidized!

STEP 6

Chlorine ( $\mathrm{Cl}$ ) goes from an oxidation state of zero to -1.
This decrease in oxidation state means $\mathrm{Cl}$ is being reduced!

STEP 7

The oxidation half-reaction shows what happens to the substance being oxidized.
In our case, it's nickel.
So, we write: $\mathrm{Ni(s)} \longrightarrow \mathrm{Ni^{2+}(aq)}$

STEP 8

The reduction half-reaction shows what happens to the substance being reduced.
Here, it's chlorine.
So, we write: $\mathrm{Cl_2(g)} \longrightarrow 2\mathrm{Cl^{-}(aq)}$ Notice how we added a 2 in front of $\mathrm{Cl^{-}}$ to balance the number of chlorine atoms on both sides.

STEP 9

STEP 10

In the reduction half-reaction, the left side has a charge of zero, and the right side has a charge of -2 (two chloride ions, each with a -1 charge).
To balance this, we add two electrons to the left side: $\mathrm{Cl_2(g)} + 2\mathrm{e^{-}} \longrightarrow 2\mathrm{Cl^{-}(aq)}$

STEP 11

The oxidation half-reaction is: $\mathrm{Ni}(\mathrm{~s}) \longrightarrow \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$ The reduction half-reaction is: $\mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}^{-}(\mathrm{aq})$

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Studdy Solution

STEP 1

STEP 2

1. Assign oxidation states2. Identify oxidation and reduction3. Write half-reactions4. Balance charges with electrons

STEP 3

Let's **assign oxidation states** to each element in the reaction.Remember, pure elements have an oxidation state of **zero**.So, Ni(s)\mathrm{Ni(s)}Ni(s) has an oxidation state of **zero**, and Cl2(g)\mathrm{Cl_2(g)}Cl2​(g) also has an oxidation state of **zero**.

STEP 4

In the products, Ni2+(aq)\mathrm{Ni^{2+}(aq)}Ni2+(aq) has an oxidation state of **+2** (it's given right there!), and each Cl−(aq)\mathrm{Cl^{-}(aq)}Cl−(aq) has an oxidation state of **-1**.

STEP 5

Now, let's see what's happening to our elements.Nickel (Ni\mathrm{Ni}Ni) goes from an oxidation state of **zero** to **+2**.This increase in oxidation state means Ni\mathrm{Ni}Ni is being **oxidized**!

STEP 6

Chlorine (Cl\mathrm{Cl}Cl) goes from an oxidation state of **zero** to **-1**.This decrease in oxidation state means Cl\mathrm{Cl}Cl is being **reduced**!

STEP 7

The **oxidation half-reaction** shows what happens to the substance being oxidized.In our case, it's nickel.So, we write: Ni(s)⟶Ni2+(aq) \mathrm{Ni(s)} \longrightarrow \mathrm{Ni^{2+}(aq)} Ni(s)⟶Ni2+(aq)

STEP 8

STEP 9

STEP 10

STEP 11

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1. Assign oxidation states
2. Identify oxidation and reduction
3. Write half-reactions
4. Balance charges with electrons

Let's assign oxidation states to each element in the reaction.
Remember, pure elements have an oxidation state of zero.
So, $\mathrm{Ni(s)}$ has an oxidation state of zero, and $\mathrm{Cl_2(g)}$ also has an oxidation state of zero.

In the products, $\mathrm{Ni^{2+}(aq)}$ has an oxidation state of +2 (it's given right there!), and each $\mathrm{Cl^{-}(aq)}$ has an oxidation state of -1.

Now, let's see what's happening to our elements.
Nickel ( $\mathrm{Ni}$ ) goes from an oxidation state of zero to +2.
This increase in oxidation state means $\mathrm{Ni}$ is being oxidized!

Chlorine ( $\mathrm{Cl}$ ) goes from an oxidation state of zero to -1.
This decrease in oxidation state means $\mathrm{Cl}$ is being reduced!

The oxidation half-reaction shows what happens to the substance being oxidized.
In our case, it's nickel.
So, we write: $\mathrm{Ni(s)} \longrightarrow \mathrm{Ni^{2+}(aq)}$