Page 397 - SUBSEC October 2017_Neat

02112020/CAPE/SPEC/MS/2017
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CHEMISTRY
UNIT 01 — PAPER 02
MARK SCHEME
KC UK XS

Question 2

S.O.: Module: 1 ----3.1, 3.2 , 3.3, 4.1, 4.2, 6.3, 6.4, 6.5

-3
2+
(a) (i) Place a strip of zinc in 1 mol dm Zn (aq) in a beaker
-3
2+
 Place a strip of copper in 1 mol dm Cu (aq) in a
beaker

 Connect the two metal strips to a high resistance
voltmeter

 Connect the two solutions by means of a salt bridge a
strip filter soaked in KNO3(aq) or KCI(aq) (or solutions
in the tube fitted with porous plugs)
 Try to maintain temperature if 25 C [5 marks] 5
0

(ii) Zn(s) Zn + 2e [1 mark]
2+
2
2+
Cu (aq) + 2e Cu(s) [1 mark]

(iii) The anode is Zn; the cathode is Cu [1 mark]
1

(iv) The electrons flow from zinc to the copper half cell
since the zinc gives up electrons more readily than
copper. [1 mark] 1 1

2+
2+
(v) Zn(s)\Zn (aq)\\Cu (aq)+ Cu(s) [1 mark]
1
(vi) Cu (aq)+ Zn(s) Zn (aq)+ Cu (s) [1 mark]
2+
2+

E cell = E Cu\Cu 2+ - E Zn\Zn 2+
θ
θ
θ
=0.34-(-0.76)=1.10V [2 marks]
(b) 2

As the reaction in (a)(v) proceeds Cu 2+ (aq) decreases while Zn 2+ (aq)

increases. Hence, according to Le Chatelier’s principle, a
(c) shift to the right will increase the cell reaction and the cell

voltage. Therefore, increasing Cu 2+ (aq)or decreasing Zn 2+ (aq)will
result in a larger voltage than 1.10V. [2 marks] 2