Redox and Electrochemistry
Redox and electrochemistry covers oxidation and reduction reactions in terms of electron transfer and oxidation states, how to construct and balance half-equations and overall redox equations, and how standard electrode potentials predict the feasibility and emf of electrochemical cells and fuel cells. Redox titrations using potassium manganate(VII) are a key required practical.
Before you start
Make sure you're comfortable with these topics first:
Method
- Assign oxidation states to each atom using the standard rules (uncombined elements = 0; simple ions = ionic charge; oxygen usually -2; hydrogen usually +1), then compare an element's oxidation state before and after reaction to identify oxidation (increase) and reduction (decrease).
- Write separate half-equations for the oxidation and reduction processes, balancing atoms first, then balancing charge by adding electrons to the more positive side.
- Combine the two half-equations by multiplying each by a suitable factor so the number of electrons lost equals the number gained, then add them together and cancel the electrons.
- To calculate a cell's standard emf, subtract the more negative standard electrode potential from the more positive one: E cell = E(positive electrode) - E(negative electrode).
- Remember the electrode with the more negative E value is the negative terminal (it is oxidised, releasing electrons into the external circuit), and a cell reaction is only feasible if E cell is positive.
- For redox titration calculations, use the balanced equation's mole ratio to link moles of the standard solution (e.g. MnO4-) to moles of the substance being analysed, then scale up to the full sample volume to find an unknown concentration or mass.
Worked example
A student determines the percentage by mass of iron in an iron(II) sulfate sample. A 1.80 g sample is dissolved in dilute sulfuric acid and made up to exactly 250 cm3 in a volumetric flask. 25.0 cm3 portions are titrated against 0.0200 mol/dm3 potassium manganate(VII) solution; the mean titre is 18.00 cm3. Equation: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 5Fe3+ + 4H2O. Calculate the percentage by mass of iron in the sample.
- Calculate moles of MnO4- used: (18.00/1000) x 0.0200 = 3.60 x 10^-4 mol.
- Use the mole ratio (1 MnO4- : 5 Fe2+) to find moles of Fe2+ in the 25.0 cm3 sample: 3.60 x 10^-4 x 5 = 1.80 x 10^-3 mol.
- Scale up to the full 250 cm3 solution (a factor of 10): 1.80 x 10^-3 x 10 = 1.80 x 10^-2 mol Fe2+ in total.
- Calculate the mass of iron: 1.80 x 10^-2 x 55.8 (Ar of Fe) = 1.0044 g.
- Calculate the percentage by mass: (1.0044/1.80) x 100 = 55.8%.
- Final answer: 55.8% (to 3 sf)
Practice questions
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Q1State the oxidation state of sulfur in the sulfate ion, SO4^2-.Show answer
Answer: +6
Q2Define the term oxidising agent.Show answer
Answer: A species that gains electrons (and is itself reduced) while causing another species to be oxidised
Q3Write the half-equation for the reduction of chlorine, Cl2, to chloride ions, Cl-.Show answer
Answer: Cl2 + 2e- -> 2Cl-
Q4Standard electrode potentials: E(Ag+/Ag) = +0.80 V, E(Fe2+/Fe) = -0.44 V. Calculate the standard emf of a cell made from these two half-cells.Show answer
Answer: +1.24 V (E cell = E(Ag) - E(Fe) = 0.80 - (-0.44) = 1.24 V)
Q5State which electrode (silver or iron) is the negative terminal in the cell described above, and explain why.Show answer
Answer: Iron; it has the more negative electrode potential, so it is oxidised more readily and releases electrons into the external circuit
Q6A 25.0 cm3 sample of iron(II) sulfate solution requires 21.40 cm3 of 0.0180 mol/dm3 potassium manganate(VII) solution to reach the end point (equation: MnO4- + 8H+ + 5Fe2+ -> Mn2+ + 5Fe3+ + 4H2O). Calculate the concentration, in mol/dm3, of the iron(II) sulfate solution.Show answer
Answer: 0.0770 mol/dm3 (moles MnO4- = 0.02140 x 0.0180 = 3.852 x 10^-4; moles Fe2+ = 5 x 3.852 x 10^-4 = 1.926 x 10^-3; concentration = 1.926 x 10^-3/0.0250 = 0.0770)
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
Dichromate(VI) ions, Cr2O7^2-, are reduced to Cr3+ in acidic solution. The half-equation is: Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. This half-equation combines with the oxidation of Sn2+ to Sn4+ (Sn2+ -> Sn4+ + 2e-) in a redox titration. Combine the two half-equations to give the overall balanced ionic equation.
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A hydrogen-oxygen fuel cell operates under acidic conditions. E(O2/H2O, acidic) = +1.23 V; E(H+/H2, acidic) = 0.00 V. Write the half-equation at each electrode and calculate the standard emf of this fuel cell.
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Standard electrode potentials: E(Fe3+/Fe2+) = +0.77 V; E(I2/I-) = +0.54 V. Predict, with a reason, whether Fe3+ ions will oxidise iodide ions, I-, to iodine, I2, under standard conditions. Write the overall ionic equation for this reaction. Explain why, even though the standard electrode potentials predict this reaction is feasible, some redox reactions predicted to be feasible are not observed to occur in practice.
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Free printable worksheet
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