QuestionJune 6, 2025

Balance the following redox reaction in acidic solution. How many protons are produced? Cr^2+(aq)+Cl_(2)(l)arrow CrO_(4)^-(aq)+Cl^-(aq) 16 5 2 10 4

Balance the following redox reaction in acidic solution. How many protons are produced? Cr^2+(aq)+Cl_(2)(l)arrow CrO_(4)^-(aq)+Cl^-(aq) 16 5 2 10 4
Balance the following redox reaction in acidic solution. How many protons are produced? Cr^2+(aq)+Cl_(2)(l)arrow CrO_(4)^-(aq)+Cl^-(aq) 16 5 2 10 4

Solution
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Answer

8 Explanation 1. Assign Oxidation States Cr^{2+} is +2, Cl_2 is 0, CrO_4^{-} is +6 for Cr, and Cl^- is -1. 2. Identify Oxidation and Reduction Cr^{2+} to CrO_4^{-} (oxidation), Cl_2 to Cl^- (reduction). 3. Write Half-Reactions Oxidation: Cr^{2+} \rightarrow CrO_4^{-} Reduction: Cl_2 \rightarrow Cl^- 4. Balance Atoms Other Than O and H Oxidation: Already balanced. Reduction: Cl_2 \rightarrow 2Cl^- 5. Balance Oxygen with Water Oxidation: Cr^{2+} \rightarrow CrO_4^{-} + 4H_2O 6. Balance Hydrogen with Protons Oxidation: Cr^{2+} + 8H^+ \rightarrow CrO_4^{-} + 4H_2O 7. Balance Charges with Electrons Oxidation: Cr^{2+} \rightarrow CrO_4^{-} + 8H^+ + 6e^- Reduction: Cl_2 + 2e^- \rightarrow 2Cl^- 8. Equalize Electron Transfer Multiply oxidation by 1 and reduction by 3: Cr^{2+} \rightarrow CrO_4^{-} + 8H^+ + 6e^- 3Cl_2 + 6e^- \rightarrow 6Cl^- 9. Combine Balanced Half-Reactions Cr^{2+} + 3Cl_2 + 8H^+ \rightarrow CrO_4^{-} + 6Cl^- + 4H_2O 10. Count Protons 8 protons are used in the balanced equation.

Explanation

1. Assign Oxidation States<br /> $Cr^{2+}$ is +2, $Cl_2$ is 0, $CrO_4^{-}$ is +6 for Cr, and $Cl^-$ is -1.<br /><br />2. Identify Oxidation and Reduction<br /> $Cr^{2+}$ to $CrO_4^{-}$ (oxidation), $Cl_2$ to $Cl^-$ (reduction).<br /><br />3. Write Half-Reactions<br /> Oxidation: $Cr^{2+} \rightarrow CrO_4^{-}$<br /> Reduction: $Cl_2 \rightarrow Cl^-$<br /><br />4. Balance Atoms Other Than O and H<br /> Oxidation: Already balanced.<br /> Reduction: $Cl_2 \rightarrow 2Cl^-$<br /><br />5. Balance Oxygen with Water<br /> Oxidation: $Cr^{2+} \rightarrow CrO_4^{-} + 4H_2O$<br /><br />6. Balance Hydrogen with Protons<br /> Oxidation: $Cr^{2+} + 8H^+ \rightarrow CrO_4^{-} + 4H_2O$<br /><br />7. Balance Charges with Electrons<br /> Oxidation: $Cr^{2+} \rightarrow CrO_4^{-} + 8H^+ + 6e^-$<br /> Reduction: $Cl_2 + 2e^- \rightarrow 2Cl^-$<br /><br />8. Equalize Electron Transfer<br /> Multiply oxidation by 1 and reduction by 3: <br /> $Cr^{2+} \rightarrow CrO_4^{-} + 8H^+ + 6e^-$<br /> $3Cl_2 + 6e^- \rightarrow 6Cl^-$<br /><br />9. Combine Balanced Half-Reactions<br /> $Cr^{2+} + 3Cl_2 + 8H^+ \rightarrow CrO_4^{-} + 6Cl^- + 4H_2O$<br /><br />10. Count Protons<br /> 8 protons are used in the balanced equation.
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