Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
(b) Cr2O72-(aq) + Cl ‾(aq) → Cr3+(aq) + Cl2(g
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C_2O_7^-2(aq) + 6Cl^-(aq) + 14H^+(aq) --> 3Cl_2(g) + 2Cr^3+(aq) + 7H_2O(l)
To balance the given oxidation-reduction reaction, we will use the half-reaction method. In this method, the reaction is split into two half-reactions, one representing the oxidation process and the other representing the reduction process. Here's how you can proceed:
Step 1: Determine the oxidation states of each element
- Chromium (Cr) has an oxidation state of +6 in Cr2O72-
- Chlorine (Cl) has an oxidation state of -1 in Cl ‾
- Chromium (Cr) has an oxidation state of +3 in Cr3+
- Chlorine (Cl) has an oxidation state of 0 in Cl2(g)
Step 2: Write the half-reactions for oxidation and reduction
Oxidation half-reaction: Cr2O72- → Cr3+
Reduction half-reaction: Cl ‾ → Cl2
Step 3: Balance the atoms other than H and O in both half-reactions
Oxidation half-reaction: Cr2O72- → 2Cr3+
Reduction half-reaction: 2Cl ‾ → Cl2
Step 4: Balance the charges in both half-reactions
Oxidation half-reaction: Cr2O72- + 14H+ → 2Cr3+ + 7H2O
Reduction half-reaction: 2Cl ‾ → Cl2 + 2e-
Step 5: Balance the electrons transferred between the half-reactions
Oxidation half-reaction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Reduction half-reaction: 2Cl ‾ → Cl2 + 2e-
Step 6: Multiply the half-reactions to make the number of electrons equal in both
Oxidation half-reaction: 2Cr2O72- + 28H+ + 12e- → 4Cr3+ + 14H2O
Reduction half-reaction: 12Cl ‾ → 6Cl2 + 12e-
Step 7: Add the two balanced half-reactions together
2Cr2O72- + 28H+ + 12e- + 12Cl ‾ → 4Cr3+ + 14H2O + 6Cl2
Finally, simplify the equation and write the states of matter:
Cr2O72-(aq) + 14H+(aq) + 6Cl-(aq) → 4Cr3+(aq) + 7H2O(l) + 3Cl2(g)
Therefore, the balanced equation for the given oxidation-reduction reaction in acidic solution is:
Cr2O72-(aq) + 14H+(aq) + 6Cl-(aq) → 4Cr3+(aq) + 7H2O(l) + 3Cl2(g)