Balance the following redox equation. hno2(aq) + h2so4(aq) + k2cr2o7(aq)= hno3(aq) + cr2(so4)3 + h20 + k2s04?
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To balance a redox equation, we need to make sure that the number of atoms and charges on each side of the equation are equal. Here's how you can balance the given equation step by step:
Step 1: Identify the atoms and their oxidation states that change during the reaction.
In this case, the atoms that change their oxidation states are N (nitrogen) and Cr (chromium). In HNO2, the oxidation state of N is +3, and in HNO3, it is +5. In K2Cr2O7, the oxidation state of Cr is +6, and in Cr2(SO4)3, it is +3.
Step 2: Write half-reactions for the oxidation and reduction processes.
Oxidation half-reaction: N (-3) -> N (+5)
Reduction half-reaction: Cr (+6) -> Cr (+3)
Step 3: Balance the atoms in each half-reaction, excluding H and O.
Oxidation half-reaction: 2HNO2(aq) -> HNO3(aq) + NO(g)
Reduction half-reaction: Cr2O7^-2(aq) -> Cr^+3(aq)
Step 4: Balance the oxygen atoms by adding H2O to the side with the fewer O atoms.
Oxidation half-reaction: 2HNO2(aq) -> HNO3(aq) + NO(g)
Reduction half-reaction: Cr2O7^-2(aq) + 14H^+(aq) -> 2Cr^+3(aq) + 7H2O(l)
Step 5: Balance the hydrogen atoms by adding H^+ to the side with the fewer H atoms.
Oxidation half-reaction: 2HNO2(aq) + H^+ -> HNO3(aq) + NO(g)
Reduction half-reaction: Cr2O7^-2(aq) + 14H^+(aq) -> 2Cr^+3(aq) + 7H2O(l)
Step 6: Balance the charge by adding electrons (e^-) to the side that needs them.
Oxidation half-reaction: 2HNO2(aq) + 2H^+ + 2e^- -> HNO3(aq) + NO(g)
Reduction half-reaction: Cr2O7^-2(aq) + 14H^+(aq) + 6e^- -> 2Cr^+3(aq) + 7H2O(l)
Step 7: Multiply the half-reactions by appropriate coefficients to make the number of electrons equal.
Oxidation half-reaction (multiplying by 6): 6HNO2(aq) + 6H^+ + 6e^- -> 6HNO3(aq) + 6NO(g)
Reduction half-reaction (multiplying by 2): 2Cr2O7^-2(aq) + 28H^+(aq) + 12e^- -> 4Cr^+3(aq) + 14H2O(l)
Step 8: Combine the half-reactions and cancel out the electrons.
Final balanced equation:
6HNO2(aq) + 6H^+ + 2Cr2O7^-2(aq) + 28H^+(aq) + 12e^- -> 6HNO3(aq) + 6NO(g) + 4Cr^+3(aq) + 14H2O(l)
Simplifying:
6HNO2(aq) + 2Cr2O7^-2(aq) + 8H^+(aq) -> 6HNO3(aq) + 6NO(g) + 4Cr^+3(aq) + 14H2O(l)
That is the balanced redox equation for the given reaction.