Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To find the balanced equation for the redox reaction involving chromium \( (Cr^{3+}) \) and chloride \( (Cl^-) \) ions, let's analyze the given half-reactions step by step and then combine them, ensuring that electrons gained and lost are balanced.
### Step-by-Step Solution:
1. Identify the Oxidation and Reduction Half-Reactions:
- Oxidation (loss of electrons):
[tex]\[ 2 Cl^- (aq) \rightarrow Cl_2(g) + 2 e^- \][/tex]
- Reduction (gain of electrons):
[tex]\[ Cr^{3+}(aq) + 3 e^- \rightarrow Cr(s) \][/tex]
2. Balance the Electrons in the Half-Reactions:
The half-reactions need to have the same number of electrons so that they can be combined. The oxidation half-reaction releases 2 electrons per chloride molecule, while the reduction half-reaction consumes 3 electrons per chromium ion.
To balance electrons, find the least common multiple of 2 and 3, which is 6. Thus:
- Multiply the oxidation half-reaction by 3:
[tex]\[ 3 \times \left[ 2 Cl^- (aq) \rightarrow Cl_2(g) + 2 e^- \right] \][/tex]
Which becomes:
[tex]\[ 6 Cl^- (aq) \rightarrow 3 Cl_2(g) + 6 e^- \][/tex]
- Multiply the reduction half-reaction by 2:
[tex]\[ 2 \times \left[ Cr^{3+}(aq) + 3 e^- \rightarrow Cr(s) \right] \][/tex]
Which becomes:
[tex]\[ 2 Cr^{3+} (aq) + 6 e^- \rightarrow 2 Cr(s) \][/tex]
3. Combine the Balanced Half-Reactions:
Now, add the two balanced half-reactions together, ensuring that the electrons on both sides cancel out:
[tex]\[ 6 Cl^- (aq) \rightarrow 3 Cl_2(g) + 6 e^- \\ 2 Cr^{3+} (aq) + 6 e^- \rightarrow 2 Cr(s) \][/tex]
Combining, we get:
[tex]\[ 2 Cr^{3+}(aq) + 6 Cl^-(aq) \rightarrow 2 Cr(s) + 3 Cl_2(g) \][/tex]
4. Double Check for Atom and Charge Balance:
- Chromium: 2 atoms on both sides
- Chlorine: 6 atoms (as \( Cl^- \)) on the reactant side and 6 atoms (as \( Cl_2 \)) on the product side
- Charges:
- Reactants: \( 2 \times (+3) + 6 \times (-1) = +6 - 6 = 0 \)
- Products: neutral atoms \( Cr \) and \( Cl_2 \)
Thus, everything is balanced!
### Final Balanced Equation:
[tex]\[ 2 Cr^{3+}(aq) + 6 Cl^-(aq) \rightarrow 2 Cr(s) + 3 Cl_2(g) \][/tex]
Hence, the balanced equation corresponds to the first option:
[tex]\[ \boxed{2 Cr ^{3+}(a q)+6 Cl ^{-}(a q) \longrightarrow 2 Cr (s)+3 Cl _2(g)} \][/tex]
### Step-by-Step Solution:
1. Identify the Oxidation and Reduction Half-Reactions:
- Oxidation (loss of electrons):
[tex]\[ 2 Cl^- (aq) \rightarrow Cl_2(g) + 2 e^- \][/tex]
- Reduction (gain of electrons):
[tex]\[ Cr^{3+}(aq) + 3 e^- \rightarrow Cr(s) \][/tex]
2. Balance the Electrons in the Half-Reactions:
The half-reactions need to have the same number of electrons so that they can be combined. The oxidation half-reaction releases 2 electrons per chloride molecule, while the reduction half-reaction consumes 3 electrons per chromium ion.
To balance electrons, find the least common multiple of 2 and 3, which is 6. Thus:
- Multiply the oxidation half-reaction by 3:
[tex]\[ 3 \times \left[ 2 Cl^- (aq) \rightarrow Cl_2(g) + 2 e^- \right] \][/tex]
Which becomes:
[tex]\[ 6 Cl^- (aq) \rightarrow 3 Cl_2(g) + 6 e^- \][/tex]
- Multiply the reduction half-reaction by 2:
[tex]\[ 2 \times \left[ Cr^{3+}(aq) + 3 e^- \rightarrow Cr(s) \right] \][/tex]
Which becomes:
[tex]\[ 2 Cr^{3+} (aq) + 6 e^- \rightarrow 2 Cr(s) \][/tex]
3. Combine the Balanced Half-Reactions:
Now, add the two balanced half-reactions together, ensuring that the electrons on both sides cancel out:
[tex]\[ 6 Cl^- (aq) \rightarrow 3 Cl_2(g) + 6 e^- \\ 2 Cr^{3+} (aq) + 6 e^- \rightarrow 2 Cr(s) \][/tex]
Combining, we get:
[tex]\[ 2 Cr^{3+}(aq) + 6 Cl^-(aq) \rightarrow 2 Cr(s) + 3 Cl_2(g) \][/tex]
4. Double Check for Atom and Charge Balance:
- Chromium: 2 atoms on both sides
- Chlorine: 6 atoms (as \( Cl^- \)) on the reactant side and 6 atoms (as \( Cl_2 \)) on the product side
- Charges:
- Reactants: \( 2 \times (+3) + 6 \times (-1) = +6 - 6 = 0 \)
- Products: neutral atoms \( Cr \) and \( Cl_2 \)
Thus, everything is balanced!
### Final Balanced Equation:
[tex]\[ 2 Cr^{3+}(aq) + 6 Cl^-(aq) \rightarrow 2 Cr(s) + 3 Cl_2(g) \][/tex]
Hence, the balanced equation corresponds to the first option:
[tex]\[ \boxed{2 Cr ^{3+}(a q)+6 Cl ^{-}(a q) \longrightarrow 2 Cr (s)+3 Cl _2(g)} \][/tex]
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.