Related Problems with Solution
Problem 1 : Balance the following redox equation occurring in acidic solution:
$$[Cr_2O_7^{2-} + HNO_2 \rightarrow Cr^{3+} + NO_3^-.]$$
Solution :
To balance this equation, follow these steps:
Step 1: Assign oxidation states to each element: $$[Cr_2O_7^{2-} : Cr^{6+}, O^{-2}]$$ $$[HNO_2 : H^{1+}, N^{3+}, O^{-2}]$$ $$[Cr^{3+} : Cr^{3+}]$$ $$[NO_3^- : N^{5+}, O^{-2}]$$
Step 2: Write down the unbalanced equation: $$[Cr_2O_7^{2-} + HNO_2 \rightarrow Cr^{3+} + NO_3^-.]$$
Step 3: Break the reaction into half-reactions for oxidation and reduction: $$[Oxidation: HNO_2 \rightarrow NO_3^-]$$ $$[Reduction: Cr_2O_7^{2-} \rightarrow Cr^{3+}]$$
Step 4: Balance each half-reaction: Oxidation: $$[HNO_2 \rightarrow NO_3^-]$$ Add 3 electrons (e⁻) to the left side to balance the charge.
Reduction: $$[Cr_2O_7^{2-} \rightarrow Cr^{3+}]$$ Add 6 electrons (e⁻) to the right side to balance the charge.
Step 5: Multiply the half-reactions by coefficients to balance the number of electrons: $$[6(HNO_2 \rightarrow NO_3^-)]$$ $$[3(Cr_2O_7^{2-} \rightarrow Cr^{3+})]$$
Step 6: Add the balanced half-reactions to get the overall balanced equation: $$[6HNO_2 + 3Cr_2O_7^{2-} \rightarrow 6NO_3^- + 3Cr^{3+}.]$$
