Math / Algebra

QuestionThe standard cell potential ( $\left.\mathrm{E}^{0}{ }_{\text {cell }}\right)$ for the voltaic cell based on the reaction below is $\qquad$ V. Cr(s) +3Fee
Half Reaction \begin{tabular}{l|c} Half Reaction & $\mathrm{E}^{\circ}(\mathrm{V})$ \\ \hline $\mathrm{Cr}^{3+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s})$ & -0.740 \\ $\mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s})$ & -0.440 \\ $\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{s})$ & +0.771 \\ $\mathrm{Sn}^{4+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq})$ & +0.154 \end{tabular} A) +1.51 B) +1.57 C) +2.99 D) +3.05 E) -1.45

Studdy Solution

STEP 1

What is this asking? We need to find the standard cell potential for a reaction involving Chromium and Iron ions. Watch out! Make sure to flip the sign of the reduction potential when reversing a reaction to make it an oxidation reaction.
Also, remember to balance the electrons transferred in the overall reaction.

STEP 2

1. Identify the half-reactions.
2. Balance the half-reactions.
3. Calculate the cell potential.

STEP 3

The overall reaction is Cr(s) + 3Fe³⁺(aq) → Cr³⁺(aq) + 3Fe²⁺(aq).
We need to find the two half-reactions that add up to this.

STEP 4

Looking at the table, we see that the two relevant half-reactions are: $Cr^{3+}(aq) + 3e^{-} \rightarrow Cr(s)$ with $E^{\circ} = -0.740 \, V$ and $Fe^{3+}(aq) + e^{-} \rightarrow Fe^{2+}(aq)$ with $E^{\circ} = +0.771 \, V$ .

STEP 5

The chromium half-reaction needs to be flipped to represent oxidation: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ .
When we flip a reaction, we flip the sign of the potential, so now $E^{\circ} = +0.740 \, V$ .
This is now our oxidation potential.

STEP 6

The iron half-reaction needs to be multiplied by 3 to balance the electrons: $3Fe^{3+}(aq) + 3e^{-} \rightarrow 3Fe^{2+}(aq)$ .
Importantly, multiplying the reaction doesn't change the potential, so $E^{\circ}$ remains +0.771 V.
This is our reduction potential.

STEP 7

Now, we add the two balanced half-reactions together: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ $3Fe^{3+}(aq) + 3e^{-} \rightarrow 3Fe^{2+}(aq)$ Adding these gives us $Cr(s) + 3Fe^{3+}(aq) + 3e^{-} \rightarrow Cr^{3+}(aq) + 3Fe^{2+}(aq) + 3e^{-}$ .

STEP 8

Notice that we have $3e^{-}$ on both sides, so we can add $-3e^{-}$ to both sides to get our balanced overall reaction: $Cr(s) + 3Fe^{3+}(aq) \rightarrow Cr^{3+}(aq) + 3Fe^{2+}(aq)$ .

STEP 9

To calculate the standard cell potential, $E^{\circ}_{cell}$ , we simply add the oxidation potential and the reduction potential: $E^{\circ}_{cell} = E^{\circ}_{oxidation} + E^{\circ}_{reduction}$ $E^{\circ}_{cell} = (+0.740 \, V) + (+0.771 \, V)$ $E^{\circ}_{cell} = +1.511 \, V$

STEP 10

The standard cell potential for the given reaction is +1.51 V (rounded to two decimal places), which corresponds to answer choice A.

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Studdy Solution

STEP 1

STEP 2

1. Identify the half-reactions.
2. Balance the half-reactions.
3. Calculate the cell potential.

STEP 3

The overall reaction is Cr(s) + 3Fe³⁺(aq) → Cr³⁺(aq) + 3Fe²⁺(aq).
We need to find the two half-reactions that add up to this.

STEP 4

Looking at the table, we see that the two relevant half-reactions are: $Cr^{3+}(aq) + 3e^{-} \rightarrow Cr(s)$ with $E^{\circ} = -0.740 \, V$ and $Fe^{3+}(aq) + e^{-} \rightarrow Fe^{2+}(aq)$ with $E^{\circ} = +0.771 \, V$ .

STEP 5

The chromium half-reaction needs to be flipped to represent oxidation: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ .
When we flip a reaction, we flip the sign of the potential, so now $E^{\circ} = +0.740 \, V$ .
This is now our oxidation potential.

STEP 6

The iron half-reaction needs to be multiplied by 3 to balance the electrons: $3Fe^{3+}(aq) + 3e^{-} \rightarrow 3Fe^{2+}(aq)$ .
Importantly, multiplying the reaction doesn't change the potential, so $E^{\circ}$ remains +0.771 V.
This is our reduction potential.

STEP 7

Now, we add the two balanced half-reactions together: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$ $3Fe^{3+}(aq) + 3e^{-} \rightarrow 3Fe^{2+}(aq)$ Adding these gives us $Cr(s) + 3Fe^{3+}(aq) + 3e^{-} \rightarrow Cr^{3+}(aq) + 3Fe^{2+}(aq) + 3e^{-}$ .

STEP 8

Notice that we have $3e^{-}$ on both sides, so we can add $-3e^{-}$ to both sides to get our balanced overall reaction: $Cr(s) + 3Fe^{3+}(aq) \rightarrow Cr^{3+}(aq) + 3Fe^{2+}(aq)$ .

STEP 9

To calculate the standard cell potential, $E^{\circ}_{cell}$ , we simply add the oxidation potential and the reduction potential: $E^{\circ}_{cell} = E^{\circ}_{oxidation} + E^{\circ}_{reduction}$ $E^{\circ}_{cell} = (+0.740 \, V) + (+0.771 \, V)$ $E^{\circ}_{cell} = +1.511 \, V$

STEP 10

The standard cell potential for the given reaction is +1.51 V (rounded to two decimal places), which corresponds to answer choice A.

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Studdy Solution

STEP 1

STEP 2

1. Identify the half-reactions.2. Balance the half-reactions.3. Calculate the cell potential.

STEP 3

The overall reaction is Cr(s) + 3Fe3+(aq) → Cr3+(aq) + 3Fe2+(aq).We need to find the two half-reactions that add up to this.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

Notice that we have 3e−3e^{-}3e− on both sides, so we can add −3e−-3e^{-}−3e− to both sides to get our balanced overall reaction: Cr(s)+3Fe3+(aq)→Cr3+(aq)+3Fe2+(aq)Cr(s) + 3Fe^{3+}(aq) \rightarrow Cr^{3+}(aq) + 3Fe^{2+}(aq)Cr(s)+3Fe3+(aq)→Cr3+(aq)+3Fe2+(aq).

STEP 9

STEP 10

The standard cell potential for the given reaction is **+1.51 V** (rounded to two decimal places), which corresponds to answer choice A.

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1. Identify the half-reactions.
2. Balance the half-reactions.
3. Calculate the cell potential.

The overall reaction is Cr(s) + 3Fe³⁺(aq) → Cr³⁺(aq) + 3Fe²⁺(aq).
We need to find the two half-reactions that add up to this.

Notice that we have $3e^{-}$ on both sides, so we can add $-3e^{-}$ to both sides to get our balanced overall reaction: $Cr(s) + 3Fe^{3+}(aq) \rightarrow Cr^{3+}(aq) + 3Fe^{2+}(aq)$ .

The standard cell potential for the given reaction is +1.51 V (rounded to two decimal places), which corresponds to answer choice A.