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Reactivity 3. What are the mechanisms of chemical change? - Electron transfer reactions

Grade 12IBChemistry

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

🔑Concepts

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Oxidation and Reduction: Oxidation is defined as the loss of electrons (e−e^{-}) and an increase in oxidation state. Reduction is defined as the gain of electrons and a decrease in oxidation state. This is often remembered by the mnemonic 'OIL RIG'.

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Oxidation States (Numbers): These are assigned to atoms to track the movement of electrons. Key rules: Elements in their natural state = 00; Monatomic ions = their charge; Oxygen is usually −2-2 (except in peroxides where it is −1-1); Hydrogen is usually +1+1 (except in metal hydrides where it is −1-1).

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Oxidizing and Reducing Agents: An oxidizing agent (oxidant) is the species that is reduced (it gains electrons and causes oxidation in another species). A reducing agent (reductant) is the species that is oxidized (it loses electrons and causes reduction in another species).

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Variable Oxidation States: Transition metals often exhibit multiple oxidation states (e.g., Fe2+Fe^{2+} and Fe3+Fe^{3+}) due to the close proximity in energy levels of the 4s4s and 3d3d orbitals.

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Balancing Redox Equations: Redox reactions can be balanced using the half-equation method. This involves balancing atoms other than HH and OO, then balancing OO using H2OH_{2}O, then HH using H+H^{+}, and finally charge using e−e^{-}.

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Voltaic (Galvanic) Cells: These convert chemical energy from spontaneous redox reactions into electrical energy. The anode (where oxidation occurs) is the negative electrode, and the cathode (where reduction occurs) is the positive electrode.

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Electrolytic Cells: These use electrical energy to drive non-spontaneous redox reactions. In these cells, the anode is positive and the cathode is negative. Electrons are pushed from the battery to the cathode.

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Standard Hydrogen Electrode (SHE): This is the reference point for all standard electrode potentials (E⊖E^{\ominus}), assigned a value of 0.00V0.00 V at 298K298 K, 100kPa100 kPa, and 1.0mol⋅dm−31.0 mol \cdot dm^{-3} H+H^{+} concentration.

📐Formulae

∑(Oxidation States in a molecule)=0\sum (\text{Oxidation States in a molecule}) = 0

∑(Oxidation States in an ion)=Charge of the ion\sum (\text{Oxidation States in an ion}) = \text{Charge of the ion}

Ecell⊖=Ereduction⊖−Eoxidation⊖E^{\ominus}_{cell} = E^{\ominus}_{reduction} - E^{\ominus}_{oxidation}

Ecell⊖=Ecathode⊖−Eanode⊖E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode}

ΔG⊖=−nFEcell⊖\Delta G^{\ominus} = -nFE^{\ominus}_{cell}

💡Examples

Problem 1:

Determine the oxidation state of Sulfur in the sulfate ion, SO42−SO_{4}^{2-}.

Solution:

Let the oxidation state of Sulfur be xx. Oxygen has an oxidation state of −2-2. The total charge of the ion is −2-2. x+4(−2)=−2x + 4(-2) = -2 x−8=−2x - 8 = -2 x=+6x = +6

Calculation for the sum: +6−8−2\begin{array}{r} +6 \\ -8 \\ \hline -2 \end{array}

Explanation:

The sum of oxidation states in a polyatomic ion must equal the net charge of that ion. By solving for the unknown, we find Sulfur is in the +6+6 state.

Problem 2:

Predict whether the reaction between Zn(s)Zn(s) and Cu2+(aq)Cu^{2+}(aq) is spontaneous given E⊖(Zn2+/Zn)=−0.76VE^{\ominus}(Zn^{2+}/Zn) = -0.76 V and E⊖(Cu2+/Cu)=+0.34VE^{\ominus}(Cu^{2+}/Cu) = +0.34 V.

Solution:

Zinc is more reactive and will undergo oxidation (anode), while Copper ions undergo reduction (cathode). Ecell⊖=Ecathode⊖−Eanode⊖E^{\ominus}_{cell} = E^{\ominus}_{cathode} - E^{\ominus}_{anode} Ecell⊖=(+0.34V)−(−0.76V)E^{\ominus}_{cell} = (+0.34 V) - (-0.76 V) Ecell⊖=+1.10VE^{\ominus}_{cell} = +1.10 V Since Ecell⊖>0E^{\ominus}_{cell} > 0, the reaction is spontaneous.

Explanation:

A positive standard cell potential indicates a spontaneous reaction under standard conditions, corresponding to a negative Gibbs free energy change (ΔG⊖<0\Delta G^{\ominus} < 0).

Problem 3:

Balance the reduction half-equation for the conversion of dichromate ions (Cr2O72−Cr_{2}O_{7}^{2-}) to chromium(III) ions (Cr3+Cr^{3+}) in acidic solution.

Solution:

  1. Balance Chromium: Cr2O72−→2Cr3+Cr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+}
  2. Balance Oxygen with H2OH_{2}O: Cr2O72−→2Cr3++7H2OCr_{2}O_{7}^{2-} \rightarrow 2Cr^{3+} + 7H_{2}O
  3. Balance Hydrogen with H+H^{+}: Cr2O72−+14H+→2Cr3++7H2OCr_{2}O_{7}^{2-} + 14H^{+} \rightarrow 2Cr^{3+} + 7H_{2}O
  4. Balance charge with e−e^{-}: Left side charge: (−2)+14(+1)=+12(-2) + 14(+1) = +12 Right side charge: 2(+3)=+62(+3) = +6 Difference is 66, so add 6e−6e^{-} to the left side. Cr2O72−+14H++6e−→2Cr3++7H2OCr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O Charge Check: +14−2−6+6\begin{array}{r} +14 \\ -2 \\ -6 \\ \hline +6 \end{array} (Left side net charge)

Explanation:

The conservation of mass and charge is required for every half-reaction. Electrons are added to the side with the more positive total charge to achieve balance.