krit.club logo

Chemical Reactions - Redox reactions (Oxidation states and identifying agents)

Grade 11A LevelChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Redox reactions involve the simultaneous processes of oxidation and reduction. These can be defined in terms of oxygen transfer, hydrogen transfer, or electron transfer.

•

Oxidation is defined as the gain of oxygen, the loss of hydrogen, or the loss of electrons (e−e^-). A mnemonic used is OIL (Oxidation Is Loss).

•

Reduction is defined as the loss of oxygen, the gain of hydrogen, or the gain of electrons (e−e^-). A mnemonic used is RIG (Reduction Is Gain).

•

Oxidation State (Oxidation Number) indicates the degree of oxidation of an atom. Rules: Atoms in elemental form (e.g., O2O_2, NaNa) are 00; Monoatomic ions equal their charge (e.g., Mg2+Mg^{2+} is +2+2); Oxygen is usually −2-2; Hydrogen is usually +1+1; the sum of oxidation states in a neutral compound is 00.

•

An Oxidising Agent is a substance that increases the oxidation state of another substance. It is itself reduced (gains electrons). Common example: Acidified Potassium Manganate(VII) (KMnO4KMnO_4), which changes from purple to colourless.

•

A Reducing Agent is a substance that decreases the oxidation state of another substance. It is itself oxidised (loses electrons). Common example: Potassium Iodide (KIKI), which changes from colourless to brown as I−I^- is oxidised to I2I_2.

📐Formulae

Oxidation half−reaction:M→Mn++ne−Oxidation \, half-reaction: M \rightarrow M^{n+} + ne^-

Reduction half−reaction:X+ne−→Xn−Reduction \, half-reaction: X + ne^- \rightarrow X^{n-}

∑(Oxidation States)compound=0\sum (Oxidation \, States)_{compound} = 0

∑(Oxidation States)polyatomic ion=Charge of ion\sum (Oxidation \, States)_{polyatomic \, ion} = Charge \, of \, ion

💡Examples

Problem 1:

Identify the oxidation states of all elements in the reaction: Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g) and identify which species is oxidised.

Solution:

Mg:0→+2Mg: 0 \rightarrow +2, H:+1→0H: +1 \rightarrow 0, Cl:−1→−1Cl: -1 \rightarrow -1.

Explanation:

MgMg is oxidised because its oxidation state increases from 00 to +2+2 (loss of electrons). HH is reduced because its oxidation state decreases from +1+1 to 00 (gain of electrons). ClCl is a spectator ion.

Problem 2:

Calculate the oxidation state of Manganese (MnMn) in the Permanganate ion MnO4−MnO_4^-.

Solution:

Let xx be the oxidation state of MnMn. Oxygen is −2-2. So, x+4(−2)=−1⇒x−8=−1⇒x=+7x + 4(-2) = -1 \Rightarrow x - 8 = -1 \Rightarrow x = +7.

Explanation:

In a polyatomic ion, the sum of oxidation states must equal the overall charge of the ion, which is −1-1 in this case.

Problem 3:

In the reaction CuO+H2→Cu+H2OCuO + H_2 \rightarrow Cu + H_2O, identify the reducing agent.

Solution:

The reducing agent is H2H_2.

Explanation:

H2H_2 gains oxygen and its oxidation state increases from 00 in H2H_2 to +1+1 in H2OH_2O. Since it undergoes oxidation, it acts as the reducing agent for CuOCuO.