
Electrochemistry
Authored by Yip Lau Kheng
Chemistry
12th Grade - University
Used 67+ times

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10 questions
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1.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
The standard electrode potentials for three half-cells are given below.
X2+ / X ; Eө = – 0.44 V
Y2+ / Y ; Eө = – 0.76 V
Z2+ / Z ; Eө = +0.34 V
Which of the following can reduce Sn2+ (aq) to Sn (s) at 298K?
(The standard electrode potential for Sn2+ (aq) / Sn (s) at 298 K is 0.14 V)
X and Y only
X, Y and Z
X2+ and Y2+only
Y2+ and Z only
2.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
What is the standard e.m.f. of the cell shown in the following diagram?
(Standard electrode potentials: Fe2+ / Fe,– 0.44V; Fe3+ / Fe,– 0.04V; Fe3+ / Fe2+ ,+ 0.77V; Pb2+ / Pb,– 0.13V)
1.67V
0.90V
0.64V
0.31V
3.
MULTIPLE SELECT QUESTION
2 mins • 1 pt
The standard electrode potentials for some half-cell reactions are given below
Half-cell reaction Eө / V
½ Br2 + e --> Br – + 1.07
Fe3+ + e --> Fe2+ + 0.77
½ I2 + e --> I – + 0.54
Which pairs of species can react with each other based on the above data?
I – and Fe3+
Br2 and Fe2+
I2 and Br –
4.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
The standard reduction electrode potential for several half-reactions at 298K are shown.
Based on the above data, which of the following statements is correct?
Fe2+ is the strongest oxidizing agent.
H2O2 can spontaneously oxidise Fe2+ ions.
The highest e.m.f. is obtained for cells I and III.
Both Fe3+ and Sn2+ ions can react with Mn2+.
5.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
After a steady current of 18.0 A was passed for 20.0 minutes through a solution of copper(II) sulphate, the mass of the impure copper anode decreased by 8.01 g. What is the percentage purity of copper in the impure copper rod? [Ar Cu = 63.5 ; Faraday’s constant = 96500 C mol-1]
73.5%
79.6%
81.2%
88.7%
6.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Which of the following is true about the Danielle cell?
The e.m.f is 1.5V
The cathode is negatively charged.
Electricity flows from the zinc electrode to the copper electrode
outside the cell.
The electron flows from anode to cathode outside the cell.
7.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Mg2+ + 2e --> Mg; E = - 2.38V
Fe3+ + e --> Fe2+; E = +0.77V
Calculate the e.m.f. of the following cell:
Mg l Mg2+ (10M) ll Fe3+ (5M) l Fe2+(1m) l Pt
- 0.86V
– 2.31V
+ 3.16V
+ 4.17v
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