Answer :
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]
