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Date May 2017 Marks available 1 Reference code 17M.3.sl.TZ2.5
Level SL Paper 3 Time zone TZ2
Command term Deduce Question number 5 Adapted from N/A

Question

Rhodium and palladium are often used together in catalytic converters. Rhodium is a good reduction catalyst whereas palladium is a good oxidation catalyst.

In a catalytic converter, carbon monoxide is converted to carbon dioxide. Outline the process for this conversion referring to the metal used.

[3]
a.

Nickel is also used as a catalyst. It is processed from an ore until nickel(II) chloride solution is obtained. Identify one metal, using sections 24 and 25 of the data booklet, which will not react with water and can be used to extract nickel from the solution.

[1]
b.i.

Deduce the redox equation for the reaction of nickel(II) chloride solution with the metal identified in (b)(i).

[1]
b.ii.

Another method of obtaining nickel is by electrolysis of a nickel(II) chloride solution. Calculate the mass of nickel, in g, obtained by passing a current of 2.50 A through the solution for exactly 1 hour. Charge (Q) = current (I) × time (t).

[2]
c.

Markscheme

carbon monoxide/CO adsorbs onto palladium/Pd

bonds stretched/weakened/broken
OR
«new» bonds formed
OR
activation energy/Ea «barrier» lowered «in both forward and reverse reactions»

products/CO2 desorb «from catalyst surface»

[3 marks]

a.

Fe/iron
OR
Zn/zinc
OR
Co/cobalt
OR
Cd/cadmium
OR
Cr/chromium

 

Accept “Mn/manganese”.

[1 mark]

 

b.i.

Ni2+(aq) + Fe(s) Ni(s) + Fe2+(aq)
OR
Ni2+(aq) + Zn(s) Ni(s) + Zn2+(aq)
OR
Ni2+(aq) + Co(s) Ni(s) + Co2+(aq)
OR
Ni2+(aq) + Cd(s) Ni(s) + Cd2+(aq)
OR
Ni2+(aq) + Cr(s) Ni(s) + Cr2+(aq)

 

Accept “3Ni2+(aq) + 2Cr(s) → 3Ni(s) + 2Cr3+(aq)”.

Do not penalize similar equations involving formation of Fe3+(aq), Mn2+(aq) OR Co3+(aq).

Ignore Cl ions.

Accept correctly balanced non-ionic equations eg, “NiCl2(aq) + Zn(s) → Ni(s) + ZnCl2(aq)” etc.

Do not allow ECF from (b)(i).

[2 mark]

b.ii.

\(n{\text{(}}{{\text{e}}^ - }{\text{)}}\) «\( = \frac{{2.50{\text{ A}} \times 3600{\text{ s}}}}{{96500{\text{ C}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\)» = 0.09326 «mol»
OR
\(n{\text{(Ni)}}\) «\( = \frac{{0.09326{\text{ mol}}}}{2}\)» = 0.04663 «mol»

\(m{\text{(Ni)}}\) «= 0.04663 mol x 58.69 g\(\,\)mol–1» = 2.74 «g»

 

Award [2] for correct final answer.

[2 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.

Syllabus sections

Core » Topic 9: Redox processes » 9.1 Oxidation and reduction
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