Transition Metal Reactions (Edexcel International A Level (IAL) Chemistry): Flashcards

Exam code: YCH11

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  • What are the four common oxidation states of vanadium, and what colour is each in aqueous solution?

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  • What are the four common oxidation states of vanadium, and what colour is each in aqueous solution?

    1. +5: yellow (VO2+)

    2. +4: blue (VO2+)

    3. +3: green (V3+)

    4. +2: violet (V2+)

  • In the reduction of vanadium from +5 to +4, the VO2+ ion is converted to .......... and the solution changes from .......... to blue.

    In the reduction of vanadium from +5 to +4, the VO2+ ion is converted to VO2+ and the solution changes from yellow to blue.

  • What is ammonium vanadate(V)?

    Ammonium vanadate(V) is a compound used to demonstrate the variable oxidation states of vanadium. When reduced by zinc under acidic conditions, it undergoes a sequence of colour changes from yellow through blue and green to violet.

  • True or False?

    Zinc can reduce vanadium all the way from +5 to +2.

    True.

    Zinc is a strong enough reducing agent to reduce vanadium through all four oxidation states from +5 (VO2+) down to +2 (V2+) in a sequence of steps.

  • Which vanadium species is the strongest oxidising agent in the sequence from +2 to +5, and why?

    VO2+ (vanadium(V)) is the strongest oxidising agent because it has the most positive standard electrode potential (+1.00 V) in the sequence.

  • Write the overall equation for the reduction of VO2+ to VO2+ by zinc under acidic conditions.

    2VO2+ (aq) + 4H+ (aq) + Zn (s) ⇌ 2VO2+ (aq) + Zn2+ (aq) + 2H2O (l)

    The reaction requires acidic conditions due to the H+ ions in the equation.

  • True or False?

    Zinc can reduce V2+ to vanadium metal.

    False.

    This reduction is not thermodynamically feasible. The standard electrode potential of the V2+/V couple (−1.18 V) is more negative than that of Zn2+/Zn (−0.76 V), so zinc is not electron-releasing with respect to V2+.

  • What is the key rule for identifying which half-equation provides the reduction reaction when predicting feasibility?

    The half-equation with the most positive (or least negative) standard electrode potential provides the reduction reaction. The other half-equation is reversed to give the oxidation reaction.

  • The formula of the vanadium(IV) oxo ion is .......... and it is coloured .......... in aqueous solution.

    The formula of the vanadium(IV) oxo ion is VO2+ and it is coloured blue in aqueous solution.

  • What colour change occurs when acid is added to a chromate solution?

    Adding acid converts yellow chromate(VI) ions (CrO42-) to orange dichromate(VI) ions (Cr2O72-).

  • Define chromate-dichromate equilibrium

    The chromate-dichromate equilibrium is the reversible interconversion between yellow CrO42- and orange Cr2O72- ions, controlled by adding acid or alkali. Both species have chromium in the +6 oxidation state.

  • True or False?

    The conversion of chromate to dichromate is a redox reaction.

    False.

    This is an acid-base reaction, not a redox reaction. Both CrO42- and Cr2O72- have chromium in the +6 oxidation state, so no change in oxidation number occurs.

  • In the reduction of Cr2O72- to Cr3+ by zinc under acidic conditions, the oxidation state of chromium changes from .......... to .......... .

    In the reduction of Cr2O72- to Cr3+ by zinc under acidic conditions, the oxidation state of chromium changes from +6 to +3.

  • What reagents and conditions are used to oxidise Cr(OH)3 to CrO42-?

    Cr(OH)3 is oxidised to CrO42- using hydrogen peroxide (H2O2) under alkaline conditions, due to the presence of OH- ions in the equation.

  • What is the colour and formula of the precipitate formed when NaOH (aq) is added to a green solution of [Cr(H2O)6]3+ (aq)?

    A grey-green precipitate of Cr(H2O)3(OH)3 (s) is formed, as three hydroxide ions deprotonate three water ligands in the complex.

  • What happens when excess NaOH (aq) is added to the Cr(H2O)3(OH)3 precipitate?

    The precipitate dissolves to give a green solution of [Cr(OH)6]3- (aq). Further deprotonation by excess OH- makes the hydroxide act as an acid, showing amphoteric behaviour.

  • True or False?

    Cr2O72- is the strongest oxidising agent in the chromium electrode potential series.

    True.

    Cr2O72- has the most positive standard electrode potential (+1.33 V) in the series, making it the strongest oxidising agent.

  • Adding alkali to a dichromate solution shifts the equilibrium towards .......... ions, causing the colour to change from .......... to yellow.

    Adding alkali to a dichromate solution shifts the equilibrium towards chromate ions, causing the colour to change from orange to yellow.

  • Which aqueous transition metal ions form a precipitate that dissolves in excess NaOH (aq)?

    Cr3+ and Zn2+ form precipitates that dissolve in excess NaOH (aq), showing amphoteric behaviour.

  • The aqueous ion [Fe(H2O)6]2+ (aq) is .......... in colour, and with limited NaOH (aq) it forms a .......... precipitate.

    The aqueous ion [Fe(H2O)6]2+ (aq) is green in colour, and with limited NaOH (aq) it forms a green precipitate.

  • What are the colour and formula of the precipitate formed when NaOH (aq) is added to Fe3+ (aq)?

    A brown precipitate of Fe(OH)3(H2O)3 (s) is formed. Fe3+ does not dissolve in excess NaOH.

  • True or False?

    Cu2+ (aq) precipitate dissolves in excess NaOH (aq) to form a dark blue solution.

    False.

    Cu2+ (aq) forms a blue precipitate with NaOH but shows no further change with excess NaOH. It is excess NH3 (aq) that dissolves the precipitate to give a dark blue solution.

  • Which ions form precipitates that dissolve in excess NH3 (aq), and what colour are the resulting solutions?

    1. Cr3+: purple solution [Cr(NH3)6]3+

    2. Co2+: yellow solution [Co(NH3)6]2+

    3. Ni2+: dark blue solution [Ni(NH3)6]2+

    4. Cu2+: dark blue solution [Cu(NH3)4(H2O)2]2+

  • Why does adding limited NH3 (aq) to a metal aqua ion produce the same precipitate as adding limited NaOH (aq)?

    Both act as bases in limited amounts, removing H+ ions from the water ligands in a deprotonation reaction. The result is the same metal hydroxide precipitate.

  • When excess NH3 (aq) is added to [Cu(H2O)6]2+ (aq), four .......... ligands replace four water molecules to form a .......... solution.

    When excess NH3 (aq) is added to [Cu(H2O)6]2+ (aq), four ammonia ligands replace four water molecules to form a dark blue solution.

  • What is the colour and formula of the precipitate formed when NaOH (aq) is added to Co2+ (aq), and does it dissolve in excess?

    A blue precipitate of Co(OH)2(H2O)4 (s) is formed. It does not dissolve in excess NaOH (aq), but does dissolve in excess NH3 (aq) to form a yellow solution.

  • True or False?

    Zn2+ (aq) forms a colourless solution and a white precipitate with NaOH.

    True.

    [Zn(H2O)6]2+ (aq) is colourless, and adding NaOH (aq) forms a white precipitate of Zn(OH)2(H2O)4 (s).

  • Define heterogeneous catalyst

    A heterogeneous catalyst is a catalyst that is in a different physical state (phase) from the reactants, typically a solid acting on gaseous or dissolved reactants.

  • What are the three steps of surface adsorption theory for heterogeneous catalysis?

    1. Adsorption: reactants attach to active sites on the catalyst surface

    2. Reaction: bonds in adsorbed molecules weaken and products form

    3. Desorption: products detach from the catalyst surface

  • True or False?

    Tungsten (W) is an ideal heterogeneous catalyst because it adsorbs reactants very strongly.

    False.

    Tungsten adsorbs reactants too strongly, preventing products from desorbing. Effective catalysts such as Ni and Pt have an intermediate adsorption strength.

  • What catalyst is used in the Contact Process to convert SO2 to SO3, and what oxidation states does it cycle through?

    Vanadium(V) oxide (V2O5) is the catalyst. It is reduced from +5 to +4 when it oxidises SO2, then re-oxidised back to +5 by oxygen.

  • In a catalytic converter, the transition metals used are .......... and .......... supported on a .......... base.

    In a catalytic converter, the transition metals used are platinum and rhodium supported on a ceramic base.

  • Write the equation for the catalytic converter reaction between NO and CO.

    2NO (g) + 2CO (g) → N2 (g) + 2CO2 (g)

    This converts two pollutants into less harmful nitrogen and carbon dioxide.

  • How do impurities reduce the effectiveness of a heterogeneous catalyst?

    Impurities can adsorb onto active sites, blocking reactants from attaching, forming strong bonds to the surface that are unlikely to desorb, and preventing bond weakening in the reactants.

  • True or False?

    Adsorption and absorption both occur at the surface of a substance.

    False.

    Adsorption occurs only at the surface of a substance, whereas absorption involves a substance becoming distributed throughout another, like water in a sponge.

  • Why is a support medium used with heterogeneous catalysts such as Rh in catalytic converters?

    A support medium maximises the surface area of the catalyst while minimising cost, as less precious metal is needed. The catalyst is spread over a hollow matrix such as a honeycomb-like ceramic structure.

  • Define homogeneous catalyst

    A homogeneous catalyst is a catalyst that is in the same physical state (phase) as the reactants, most commonly in aqueous solution. A key feature is the formation of an intermediate species with a specific formula.

  • Why is the reaction between S2O82- and I- slow despite being energetically favourable?

    Both S2O82- and I- are negatively charged, so electrostatic repulsion between them reduces the frequency of successful collisions.

  • How do Fe2+ ions catalyse the reaction between peroxodisulfate ions and iodide ions?

    1. Fe2+ reduces S2O82- to SO42-, forming Fe3+

    2. Fe3+ then oxidises I- to I2, reforming Fe2+

    The iron cycles between +2 and +3, acting as both reducing and oxidising agent.

  • In the first step of Fe2+ catalysis, S2O82- is reduced to .......... and the iron is oxidised to .......... .

    In the first step of Fe2+ catalysis, S2O82- is reduced to SO42- and the iron is oxidised to Fe3+.

  • True or False?

    Homogeneous catalysts form an intermediate species that can be assigned a specific chemical formula.

    True.

    The formation of an intermediate with a defined formula is the key feature that distinguishes homogeneous catalysis from heterogeneous catalysis.

  • What observation would confirm that iodine has been produced in the Fe2+-catalysed reaction of S2O82- with I-?

    Adding starch to the reaction mixture, which forms a blue-black colour in the presence of iodine.

  • Why can transition metal ions such as Fe2+ act as homogeneous catalysts in redox reactions?

    They can adopt more than one stable oxidation state, allowing them to accept and donate electrons, acting as both oxidising and reducing agents in a catalytic cycle.

  • True or False?

    In the Fe2+-catalysed peroxodisulfate-iodide reaction, all species are in the aqueous phase.

    True.

    The reactants (S2O82-, I-), the catalyst (Fe2+/Fe3+), and the products are all in aqueous solution, confirming this is homogeneous catalysis.

  • Define autocatalysis

    Autocatalysis is a process in which a product of the reaction acts as a catalyst, causing the reaction rate to increase as more product accumulates over time.

  • What is the unusual shape of a concentration-time graph for an autocatalytic reaction, and why?

    The gradient becomes steeper over time, meaning the rate speeds up rather than slowing down. This is because the product that acts as the catalyst accumulates, accelerating the reaction.

  • What is the autocatalyst in the reaction between manganate(VII) ions and oxalate ions?

    Mn2+ (manganese(II) ions) are the autocatalyst. They are produced as a product of the reaction and then speed up further reaction by cycling between the +2 and +3 oxidation states.

  • In autocatalysis with MnO4- and C2O42-, the Mn2+ formed initially reacts with .......... ions to form Mn3+, which then oxidises .......... to regenerate Mn2+.

    In autocatalysis with MnO4- and C2O42-, the Mn2+ formed initially reacts with MnO4- ions to form Mn3+, which then oxidises C2O42- to regenerate Mn2+.

  • True or False?

    Mn2+ is present at the start of the MnO4-/C2O42- reaction.

    False.

    Mn2+ is not present at the beginning; it is generated as the reaction proceeds. Once formed, it accelerates the reaction as an autocatalyst.

  • How can the autocatalytic reaction between MnO4- and C2O42- be monitored experimentally?

    A colorimeter can be used to track the rate at which the purple manganate(VII) colour is consumed, which accelerates over time as Mn2+ accumulates.

  • What oxidation state cycle does Mn2+ undergo when acting as an autocatalyst in the manganate(VII)-oxalate reaction?

    Mn2+ is oxidised to Mn3+ in the first step, then Mn3+ is reduced back to Mn2+ in the second step, cycling between the +2 and +3 oxidation states.

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