75 free MCQs on Redox Reactions, each with its own worked answer and explanation. Master electron transfer in chemistry: assign oxidation states, balance half-reactions, identify oxidising and reducing agents, and connect redox to everyday reactions.
75 practice questions on Redox Reactions, sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the Redox Reactions notes.
Electron transfer in a redox reaction: zinc loses electrons (oxidised, acts as reducing agent) and copper(II) ions gain those same electrons (reduced, acts as oxidising agent).
Easy - 25 questions
Q1.
The oxidation number of oxygen in oxygen difluoride (OF<sub>2</sub>) is:
In the reaction: Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> + Fe<sup>2+</sup> → Cr<sup>3+</sup> + Fe<sup>3+</sup> (acidic), how many Fe<sup>2+</sup> ions are oxidised per one Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> ion?
In the ion-electron method of balancing, when balancing the following half-reaction in acidic medium: Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> → Cr<sup>3+</sup>, how many electrons are transferred?
The method of balancing redox reactions that involves calculating the change in oxidation numbers and multiplying coefficients to make total increase = total decrease is called:
In the balanced acidic-medium reaction MnO<sub>4</sub><sup>−</sup> + Fe<sup>2+</sup> + H<sup>+</sup> → Mn<sup>2+</sup> + Fe<sup>3+</sup> + H<sub>2</sub>O, the mole ratio of MnO<sub>4</sub><sup>−</sup> to Fe<sup>2+</sup> is:
Balance the following redox reaction in acidic medium: MnO<sub>4</sub><sup>-</sup> + C<sub>2</sub>O<sub>4</sub><sup>2-</sup> → Mn<sup>2+</sup> + CO<sub>2</sub>. What are the stoichiometric coefficients (MnO<sub>4</sub><sup>-</sup> : C<sub>2</sub>O<sub>4</sub><sup>2-</sup>)?
In the reaction: I<sub>2</sub> + 2S<sub>2</sub>O<sub>3</sub><sup>2-</sup> → 2I- + S<sub>4</sub>O<sub>6</sub><sup>2-</sup>, the oxidation state of sulphur in S<sub>4</sub>O<sub>6</sub><sup>2-</sup> is:
The comproportionation reaction (reverse of disproportionation) has the general form: A(higher OS) + A(lower OS) → A(intermediate OS). Which of the following is an example?
A 2H<sub>2</sub>O<sub>2</sub> → 2H<sub>2</sub>O + O<sub>2</sub>
B Cu<sup>2+</sup> + Cu → 2Cu+
C Cl<sub>2</sub> + 2NaOH → NaCl + NaOCl + H<sub>2</sub>O
D 2MnO<sub>4</sub><sup>-</sup> + 5H<sub>2</sub>C<sub>2</sub>O<sub>4</sub> → 2Mn<sup>2+</sup> + 10CO<sub>2</sub> + 8H<sub>2</sub>O
Balance the redox reaction in basic medium: Cr<sup>3+</sup> + H<sub>2</sub>O<sub>2</sub> → CrO<sub>4</sub><sup>2-</sup> + H<sub>2</sub>O. What is the oxidising agent?
A Cr<sup>3+</sup>, the species that is itself oxidised to chromate in this reaction
B H<sub>2</sub>O<sub>2</sub> (H<sub>2</sub>O<sub>2</sub> oxidises Cr<sup>3+</sup> to CrO<sub>4</sub><sup>2-</sup>)
C H<sub>2</sub>O, the product formed and therefore not the oxidising agent
D OH-, the basic medium ion that does not change oxidation state
In calculating n-factor for Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> in its reaction with I<sub>2</sub> (where products include Na<sub>2</sub>S<sub>4</sub>O<sub>6</sub>), the n-factor of Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is:
During iodometric back-titration, Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is used to titrate liberated I<sub>2</sub>. The reaction is: 2S<sub>2</sub>O<sub>3</sub><sup>2-</sup> + I<sub>2</sub> → S<sub>4</sub>O<sub>6</sub><sup>2-</sup> + 2I-. What is the role of Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> here?
A sample of iron ore weighing 1.00 g is dissolved in acid and all iron converted to Fe<sup>2+</sup>. The solution requires 25.0 mL of 0.0200 M KMnO<sub>4</sub> for titration. The percentage of Fe in the ore is approximately:
Using the standard electrode potentials E°(Fe<sup>3+</sup>/Fe<sup>2+</sup>) = +0.77 V and E°(I<sub>2</sub>/I-) = +0.54 V, predict whether Fe<sup>3+</sup> will oxidise I- ions:
A No, generally because E°(Fe<sup>3+</sup>/Fe<sup>2+</sup>) is said to be less than E°(I<sub>2</sub>/I-) under these conditions
B Yes, because E°(Fe<sup>3+</sup>/Fe<sup>2+</sup>) > E°(I<sub>2</sub>/I-) so the cell EMF is positive and the reaction is spontaneous
C No reaction occurs, since both given half-reaction potentials are positive values in routine practice
D Both species are said to function mainly as oxidising agents and so cannot react together overall