Q.Balance the following ionic equation in acidic medium by the oxidation-number-change method: $\text{MnO}_4^{-} + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$.

Balancing by the oxidation-number method: identify the atoms that change oxidation number, make total electrons lost equal total electrons gained via coefficients, then balance the rest of the atoms.

1. KMnO4 + Na2SO3 -> MnO2 + Na2SO4 + KOH Step 1 -- Identify oxidation number changes: Mn: +7 (in KMnO4) -> +4 (in MnO2); change = -3 (gain of 3 electrons, reduction). S: +4 (in Na2SO3) -> +6 (in Na2SO4); change = +2 (loss of 2 electrons, oxidation). Step 2 -- Equalise electrons: LCM of 3 and 2 is 6, so use 2 Mn (2x3=6 e gained) and 3 S (3x2=6 e lost): 2 KMnO4 + 3 Na2SO3 -> 2 MnO2 + 3 Na2SO4 Step 3 -- Balance remaining atoms: K: 2 on left -> need 2 KOH on right. Na: 3x2 = 6 on left -> 3 Na2SO4 gives 6 Na on right (balanced). Balance O and H by adding H2O: left O so far = 2(4)+3(3) = 17; right O so far = 2(2)+3(4)+2(1, from 2KOH) = 4+12+2 = 18. Adding 1 H2O to the left supplies 1 more O (giving 18, balanced) and 2 H, which appear as the 2 H in 2 KOH on the right. Final balanced equation: 2 KMnO4 + 3 Na2SO3 + H2O -> 2 MnO2 + 3 Na2SO4 + 2 KOH
2. Cu + HNO3 -> Cu(NO3)2 + NO2 + H2O Step 1 -- Identify oxidation number changes: Cu: 0 (metal) -> +2 (in Cu(NO3)2); change = +2 (loss of 2 electrons, oxidation). N: +5 (in HNO3) -> +4 (in NO2); change = -1 (gain of 1 electron, reduction) -- this applies only to the HNO3 that gets reduced to NO2. Step 2 -- Equalise electrons: Cu loses 2 electrons, so 2 nitrogens must each gain 1 electron -> 2 HNO3 act as the oxidising agent (reduced to 2 NO2). …