Chemistry Multiple Choice Questions on “Electrochemistry – Galvanic Cells”.
1. A galvanic cell converts electrical energy into chemical energy.
a) True
b) False
Answer: b
Clarification: A galvanic cell is a type of electrochemical cell that converts chemical energy into electrical energy. The electrochemical cell which converts electrical energy into chemical energy is called electrolytic cell.
2. Who invented the galvanic cell?
a) Galvani and Volta
b) Henry Cavendish
c) Joseph Priestley
d) Antoine Lavoisier
Answer: a
Clarification: Electrochemical cells are also called galvanic or voltaic cells, after the names of Luigi Galvani and Alessandro Volta who were the first to perform experiments on the conversion of chemical energy into electrical energy.
3. Which of the following electrolytes is not preferred in a salt bridge?
a) KCl
b) KNO3
c) NH4NO3
d) NaCl
Answer: d
Clarification: In a salt bridge, the electrolytes like KCl, KNO3 or NH4NO3 are preferred because their ions have almost equal transport number, viz., 0.5, i.e., they move with almost the same speed when an electric current flows through them.
4. Which of the following is false regarding galvanic cells?
a) It converts chemical energy into electrical energy
b) The electrolytes taken in the two beakers are different
c) The reactions taking place are non-spontaneous
d) To set up this cell, a salt bridge is used
Answer: c
Clarification: Galvanic cells are used to convert chemical energy into electrical energy. Two electrodes are usually set up in two separate beakers. The electrolytes taken in the two beakers are different. Galvanic cells are based upon spontaneous redox reactions. A salt bridge is used to set up this cell.
5. The electrode on which oxidation occurs is called the anode. True or False?
a) True
b) False
Answer: a
Clarification: An anode is an electrode where oxidation takes place. An anode is a negative pole in a galvanic cell. In an electrolytic cell, the anode acts as the positive pole. Cathodes are electrodes where reduction takes place.
6. A cell is prepared by dipping a copper rod in 1 M CuSO4 solution and an iron rod in 2 M FeSO4 solution. What are the cathode and anode respectively?
a) Cathode: Iron, Anode: Copper
b) Cathode: Copper, Anode: Iron
c) Cathode: Iron, Anode: Iron
d) Cathode: Copper, Anode: Copper
Answer: b
Clarification: The given cell is represented as:
Fe (s) | FeSO4 (2 M) || CuSO4 (1 M) | Cu (s)
Since the E° of iron
7. Which of the following is the correct order of reactivity of metals?
a) Zn > Mg > Fe > Cu > Ag
b) Zn > Mg > Fe > Ag > Cu
c) Mg > Zn > Fe > Ag > Cu
d) Mg > Zn > Fe > Cu > Ag
View Answer
Answer: d
Clarification: Greater the oxidation potential of metal, the more easily it can lose electrons and hence greater is its reactivity. As a result, a metal with greater oxidation potential can displace metals with lower oxidation potentials from their salt solutions. Hence, the correct order of reactivity is Mg > Zn > Fe > Cu > Ag.
8. Which of the following is a correct method to calculate the EMF of a galvanic cell?
a) Standard EMF of the cell = [Standard reduction potential of the reduction half reaction] + [Standard reduction potential of the oxidation half reaction]
b) Standard EMF of the cell = [Standard oxidation potential of the oxidation half reaction] – [Standard reduction potential of the reduction half reaction]
c) E°cell = E°cathode – E°anode
d) Standard EMF of the cell = [Standard reduction potential of the right hand side electrode] + [Standard reduction potential of the left hand side electrode]
Answer: c
Clarification: The correct methods to calculate the EMF of a galvanic cell are:
Standard EMF of the cell = [Standard reduction potential of the reduction half reaction] – [Standard reduction potential of the oxidation half reaction].
Standard EMF of the cell = [Standard oxidation potential of the oxidation half reaction] + [Standard reduction potential of the reduction half reaction].
E°cell = E°cathode – E°anode.
Standard EMF of the cell = [Standard reduction potential of the right hand side electrode] – [Standard reduction potential of the left hand side electrode].
9. What is the EMF of a galvanic cell if E°cathode = 0.80 volts and E°anode = -0.76 volts?
a) 1.56 volts
b) 0.04 volts
c) -1.56 volts
d) -0.04 volts
Answer: a
Clarification: Given,
E°cathode = 0.80 volts
E°anode = -0.76 volts
E°cell = E°cathode – E°anode
E°cell = 0.80 – (-0.76)
E°cell = 1.56 volts.
10. What is the EMF of a galvanic cell if the standard oxidation potential of the oxidation half-reaction is 0.64 volts and the standard reduction potential of the reduction half-reaction is 0.48 volts?
a) 1.48 volts
b) 1.12 volts
c) 1.36 volts
d) 0.96 volts
Answer: b
Clarification: Given,
Standard oxidation potential of the oxidation half reaction = 0.64 volts
Standard reduction potential of the reduction half reaction = 0.48 volts
Standard EMF of the cell = [Standard oxidation potential of the oxidation half reaction] + [Standard reduction potential of the reduction half reaction]
= 0.64 + 0.48
= 1.12 volts.
11. What is the EMF of a galvanic cell if the standard reduction potential of the reduction half-reaction is -0.38 volts and the standard reduction potential of the oxidation half-reaction is 0.52 volts?
a) -0.9 volts
b) -0.6 volts
c) 0.9 volts
d) 0.6 volts
Answer: a
Clarification: Given,
Standard reduction potential of the reduction half reaction = -0.38 volts
Standard reduction potential of the oxidation half reaction = 0.52 volts
Standard EMF of the cell = [Standard reduction potential of the reduction half reaction] – [Standard reduction potential of the oxidation half reaction]
= -0.38 – (0.52)
= -0.9 volts.
12. What is the standard reduction potential of the cathode of a galvanic cell if the standard EMF of the cell and the standard reduction potential of the anode are 2.71 and -2.37 respectively?
a) 0.68 volts
b) -0.68 volts
c) -0.34 volts
d) 0.34 volts
Answer: d
Clarification: Given,
Standard EMF of the cell = E°cell = 2.71 volts
Standard reduction potential of the anode = E°anode = -2.37 volts
E°cell = E°cathode – E°anode
E°cathode = E°cell + E°anode
= 2.71 + (-2.37)
= 0.34 volts.