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NEET Chemistry MCQ
Redox Reactions And Electro Chemistry Mcq
Quiz 8
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Q.1
For measuring the conductivity of an electrolyte, its solution should be prepared is [ Bihar CEE 1990]
0%
a) tap water
0%
b) distilled water
0%
c) conductivity water
0%
d) polywater
Explanation
Conductivity water has negligible conductance of its own Answer:(c)
Q.2
In an electroplating experiment m gm silver is deposited when 4 amp current flows for 2 minutes. The amount (in gram) of silver deposited by 6ampof current flowing for 40sec will be [ MLNR 1991]
0%
a) 4m
0%
b) m/2
0%
c) 2m
0%
d) m/4
Explanation
From first law m=Zit ∴ Z=m/(4×2×60)=m/ 480 Now m'=(m/480) × 6 ×40=m/2 Answer: (b)
Q.3
One faraday is equal to ..[ AFMC 1989]
0%
a)69500 coulombs
0%
b) 99500 coulombs
0%
c)96500 coulombs approximately
0%
d)none of the above
Explanation
Answer: (c)
Q.4
NaOH is manufactured by the electrolysis of brine solution. the products of the reaction are ..[ KCET 1990]
0%
a) Cl2 and H2
0%
b) Cl2 and Na-Hg
0%
c)Cl2 and Na
0%
d)Cl2 and O2
Explanation
Answer: (a)
Q.5
A current of strength 2.5 amp was passed through CuSO4 solution for 6 minutes 26 seconds. the amount of copper deposited is ( At. Wt of Cu=63.5) [ EAMCET 1989]
0%
a) 0.3175 g
0%
b) 3.175g
0%
c)0.635 g
0%
d)6.35g
Explanation
Total charge=2.5×(6×60 +26)=965Charge in faraday=965/96500=0.01 F Now 1 F charge deposit 1 gram equivalent of Cu=63.5/2=31.75 grams∴0.01 F will deposit 0.01×31.75=0.3175g of Cu Answer:(a)
Q.6
The amount of charge that must be passed through a solution containing Cu2+ in order to deposited 1gm atom of copper (65.5) is .. [ EAMCET 1991]
0%
a) 1520 Coulomb
0%
b) 3040 coulomb
0%
c) 96500 coulomb
0%
d) 193000 coulomb
Explanation
Equivalent weight of Cu=63.5 / 2 Thus 2F charge can deposit 63.5 gm of Cu Now 1 gram atom=63.5gm given deposited Thus 2F charge=2×96500=193000 coulomb charge Answer: (d)
Q.7
2H+(aq) + 2e- → H2(g) Standard electrode potential for the above reaction ( in volts ) is [ CPMT 1988]
0%
a)0
0%
b) +1
0%
c)-1
0%
d)None of these
Explanation
Answer: (a)
Q.8
The amount of electricity required to deposit 1 mole of aluminum from a solution of AlCl3 would be ..[ haryana CEET 1991]
0%
a) 0.33 Faraday
0%
b) 1 Faraday
0%
c)3 Faraday
0%
d)1 Ampere
Explanation
Valancy of Al is one thus three mole of electricity required to deposit one mole of Al 3 mole of electricity=3 FaradayAnswer: (c)
Q.9
Select the strongest reducing agent from amongst the substances involved in the following half-reactions.Cl2(g) + e- → 2Cl-(aq) +1.36 VBr2+e- → 2Br- (aq) +1.09VSn2+ +2e- → Sn -0.14V[Pb. CET 1989]
0%
a) Cl-
0%
b) Br-
0%
c)Sn
0%
d)Sn2+
Explanation
More the reduction potential stronger reducing agent. ADSFSHKJHKLJHL:JLJLjl Answer:(c)
Q.10
The electrochemical equivalent of silver is 0.001118 g. When an electric current of 0.5 ampere is passed through an aqueous solver nitrate solution for 200 seconds, the amount of silver deposited is [ BHU 1987]
0%
a) 1.118 g
0%
b) 0.1118 g
0%
c) 5.590 g
0%
d) 0.5590 g
Explanation
m=ZIt m=(0.00118) × (0.5) × (200)=0.1118gm Answer: (b)
Q.11
The cathodic reaction in electrolysis of dilute H2SO4 with platinum electrodes is [ MLNR 1988]
0%
a)Oxidation
0%
b) reduction
0%
c)Both
0%
d)Neutralization
Explanation
Answer: (b)
Q.12
When an electric current is passed through acidulated water, 112 ml of hydrogen gas at NTP is collected at the cathode in 965 second. the current passed in amperes is ..[ MLNR 1991]
0%
a) 1.0
0%
b)0.5
0%
c)0.1
0%
d)2.0
Explanation
112ml of H2=22400 cc / 200 Volume of one mole of hydrogen gas at NTP=22400 ccThus 112 ml=1/200 moles of H2Gram equivalence of H2=2 Thus One mole of charge (F) liberates 0.5 moles of H2∴ To liberate 1/200 moles=1/100 moles of charge required 1/100 moles of charge=F/100=96500/100=965 C Now I=Q/t=965/965=1Amp Answer: (a)
Q.13
Given electrode potentials: Ag+/Ag=0.80V, Co2+/Co=0.28VCu2+/Cu=0.34V, Zn2+/Zn=-0.76VThe most reactive metal which displaces other metals from their salts in solution is [ Pb. CET 1990]
0%
a) Ag
0%
b) Cu
0%
c)Co
0%
d)Zn
Explanation
reduction potential of Zn is more than other metals Answer:(d)
Q.14
How many coulombs of electricity are consumed when 100 mA current is passed through a solution of AgNO3 for half an hour during an electrolysis experiment? [ Pb CET 1989]
0%
a) 108
0%
b) 180
0%
c) 1800
0%
d) 18000
Explanation
Electricity Q - It Q=(100 × 10-3) × 1800 Q=180 Answer: (b)
Q.15
The most powerful reducing agent is[ MLNR 1989]
0%
a)Mg
0%
b) K
0%
c)Na
0%
d)Ba
Explanation
K have more negative reduction potential thus more powerful reducing agentAnswer: (b)
Q.16
The electrolytic sells, one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. the ratio of iron deposited at cathodes in the two cells when electricity is passed through the cell will be [ CPMT 1989]
0%
a) 3:1
0%
b) 2:1
0%
c)1:1
0%
d)3:2
Explanation
Formula for ferrous chloride is FeCl2 Equivalent weight of Fe2+=56/2 Formula for ferric chloride if FeCl3 Equivalent weight of Fe3+=56/3 Ratio of deposit is Fe2+ : Fe3+=(56/2) : (56/3)=3:2Answer: (d)
Q.17
Three faraday electricity was passed through an aqueous solution of iron(II) bromide. the weight of iron metal (At wt=56) deposited at the cathode ( in gm) is [ EAMCET 1991]
0%
a) 56
0%
b) 84
0%
c)112
0%
d)168
Explanation
Gram equivalence of Fe(II)=56/2=28 Thus One faraday can deposit=28 gm of Fe 3 F charge will deposit=3×28=84 gm Answer:(b)
Q.18
Of the following metals that cannot be obtained by electrolysis of the aqueous solution of their salts are [ IIT 1990]
0%
a) Ag and Mg
0%
b) Ag and Al
0%
c) Mg and Al
0%
d) Cu and Cr
Explanation
Mg and Al have lower reduction potential than H2O. Hence H2O is reduced more easily to give H2 gas at the cathode Answer: (c)
Q.19
Electrolytic cell is used to convert .. [ AFMC 1989]
0%
a)chemical energy to electrical energy
0%
b) electrical energy to chemical energy
0%
c)chemical energy to mechanical energy
0%
d)electrical energy to mechanical energy
Explanation
Answer: (b)
Q.20
The charge required to liberate 11.5 g sodium from fused sodium chloride is [ AIIMS 1992]
0%
a) 0.5 faraday
0%
b) 1.0 faraday
0%
c)1.5 faraday
0%
d)96500 coulomb
Explanation
Atomic weigh of sodium=23 Gram equivalence of Na=23/1=23Thus 0.5 F charge required to liberate=11.5 g of Na Answer: (a)
Q.21
K, Ca and Li metals may be arranged in the decreasing order of their standard electrode potential as [ CPMT 1990]
0%
a) K, Ca, Li
0%
b) Li, K, Ca
0%
c)Li, Ca, K
0%
d)Ca, K, Li
Explanation
Answer:(b)
Q.22
Electrolysis of molten NaCl leads to the formation of [ KCET 1990]
0%
a) Sodium and Hydrogen
0%
b) Sodium and Oxygen
0%
c) Hydrogen and Oxygen
0%
d) Sodium and Chlorine
Explanation
Answer: (d)
Q.23
On the basis of position in the electrochemical series, the metal which does not displace hydrogen from water and acid is [ MP PMT 1988]
0%
a)Hg
0%
b) Al
0%
c)Pb
0%
d)Ba
Explanation
Hg has lower oxidation potential than HydrogenAnswer: (a)
Q.24
A certain current deposit 0.50 of hydrogen in 2 hrs. The amount of copper liberated from a solution of copper sulphate by the same current flowing for the same time would be.. [ CPMT 1991]
0%
a) 31.8 g
0%
b) 63.6 g
0%
c)15.9 g
0%
d)6.36g
Explanation
0.5 gm=1/2 moles of HydrogenGram equivalent of H=1 gm Since 0.5 moles of Hydrogen liberated=0.5gram mole liberatedThus 0.5F current is used Now gram equivalence of Cu=63.5/2=31.8 gCupper liberated by 0.5F=0.5 × 31.8=15.9 g Answer: (c)
Q.25
The emf of the following three galvanic cells I) Zn|Zn2+(1M)|| Cu2+(1M)|CuII) Zn|Zn2+(0.1M)|| Cu2+(1M)|CuIII) Zn|Zn2+(1m)|| Cu2+(0.1M)|Cu are represented by E1, E2, EWhich of the following statement is true? [ Pb CET 1990]
0%
a) E1 > E2 > E3
0%
b)E3 > E2 > E1
0%
c)E3 > E1 > E2
0%
d) E1 > E3 > E2
Explanation
According to Nernst equation crates smaller the [Zn2+] / Cu2+ ratio greater the cell potential Answer:(d)
Q.26
The quantity of electricity needed to liberated 0.5 gram equivalent of an element is [ CPMT 1988]
0%
a) 48250 faraday
0%
b) 48250 coulomb
0%
c) 19300 faraday
0%
d) 19300 coulomb
Explanation
1Faraday=9500 coulomb charge liberate 1 gram equivalent substance Answer: (b)
Q.27
Eo value of Mg2+/Mg is -2.37V of Zn2+/Zn is -0.76V and Fe2+/Fe is -0.44V. Which of the following statement is correct? [ EAMCET 1989]
0%
a)Zn will reduce Fe2+
0%
b) Zn will reduce Mg2+
0%
c)Mg oxidises Fe
0%
d)Zn oxidizes Fe
Explanation
More the reduction potential better is the reducing agentAnswer: (a)
Q.28
When 9650 coulombs of electricity is passed through a solution of copper sulphate, the amount of copper deposited is [ At Wt of Cu=63.6) [ Pb CET 1990]
0%
a) 0.318 g
0%
b) 3.18 g
0%
c)31.8g
0%
d)63.6 g
Explanation
9650 coulomb=0.1 F charge Cu deposited=0.1 gram equivalent 1 gram equivalent of copper=63.6/2 Now 0.1(63.6/2)=3.18 gAnswer: (b)
Q.29
During electrolysis of NaCl solution, the gas liberated at the anode is [ Pb CET 1990]
0%
a) H2
0%
b) O2
0%
c)Cl2
0%
d)None of these
Explanation
Answer:(c)
Q.30
To deposited 0.6354 g of copper by electrolysis of aqueous cupric sulphate solution
0%
a) 9650
0%
b) 4825
0%
c) 3860
0%
d) 1930
Explanation
2F Charge required to deposited one mole of Cu=63.54 gm This to deposit 0.6354 gm Charge required is=2F/100=(2×96500)/100=1930 Answer: (d)
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