Tag: chemistry

Questions Related to chemistry

In an electrochemical cell, anode and cathode are__________.

electrical cells patterns and properties of metals chemistry

  1. Positively and negatively charges ions

  2. Positively and negatively charges electrodes

  3. Negatively and positively charged electrodes

  4. Negatively and positively charged ions

Show answer
Correct Option: B
Explanation:

In both kinds of electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs. The anode is considered positive and the cathode is considered negative. 


Hence, the correct option is $\text{B}$

Electrochemical cells are also called:

electrical cells patterns and properties of metals chemistry

  1. electrolytic cells

  2. galvanic cells

  3. fuel cells

  4. voltaic cells

Show answer
Correct Option: A,B,D
Explanation:

Electrochemical cell which convert electrical energy into chemical energy is called as electrolytic cell. 
Electrochemical cell which converts chemical energy into electrical energy is called as Galvanic or Voltaic cell

Foe the electrochemical cell:
$Zn\left( s \right) |{ Zn }^{ 2+ }\left( aq \right) \parallel { Cl }^{ - }\left( aq \right) |{ Cl } _{ 2 }\left( g \right) |Pt\left( s \right) $
Given : ${ E } _{ { Zn }^{ 2+ }/Zn }^{ o }=-0.76\ Volt$
             ${ E } _{ { Cl }^{ - }/{ Cl } _{ 2 }\left( g \right)  }^{ o }=-1.36\ Volt$
From these data one can deduce that:

electrical cells patterns and properties of metals chemistry

  1. $Zn + Cl _{2} \rightleftharpoons Zn^{2+} + 2Cl^{-}$ is a non-spontaneous reaction at standard conditions.

  2. $Zn^{2+} + 2Cl^{-} \rightleftharpoons Cl _{2} + Zn$ is a spontaneous reaction at standard conditions with ${ E } _{ cell }^{ o } = 2.12\ volt.$

  3. $Zn + Cl _{2} \longrightarrow Zn^{2+} + 2 Cl^{-}$ is a spontaneous reaction at standard conditions with ${ E } _{ cell }^{ o } = 2.12\ volt.$

  4. $Zn + Cl _{2} \longrightarrow Zn^{2+} + 2 Cl^{-}$ is a spontaneous reaction at standard conditions with ${ E } _{ cell }^{ o } = 0.60\ volt.$

Show answer
Correct Option: C
Explanation:
$Cl=\to zn^{2+} +2e^- \to Zn$
$2Cl\to Cl _2^- +2e^-$
for sponlareous reaction $E^o$ cell $>$ zero
$\Delta G <$ zero
$Zn^{3+}+2Cl^{-}\to Cl _2+Zn$ 
$Zn \to Zn^{2+} +2e^-$
$Cl _2 \to 2e^- \to 2Cl^- $(cathode)
$Zn +Cl _2 \to Zn^{3+}+2Cl^- ....(1)$
$E^o _{RP}$ (cathode) $=0.76$
$E^o $ cell $=E^o _{OP}+E^o _{RP}$
$=2.12$
as $E^o > 0$
$\Delta G < $ zero
so reaction $(1)$ is sponlareous at $2.12$

In an electrolytic cell:

electrical cells patterns and properties of metals chemistry

  1. anode is positively charged

  2. cathode is negatively charged

  3. oxidation takes place at anode

  4. reduction takes place at cathode

Show answer
Correct Option: A,B,C,D
Explanation:

In electrolytic cell,

anode is positively charged and $-ve$ charged ions are attracted to it and oxidation takes place by loss of electron,
cathode is negatively charged and $+ve$ charge ions are attracted to it and reduction takes place by gain of electron;
hence, all options are corrrect. 

The electrochemical cell shown below is a concentration cell.
$M|{ M }^{ 2+ }$ (saturated solution of a sparingly soluble salt, $M{X} _{2})\parallel {M}^{2+}(0.001 mol{dm}^{-3})| M$. 
The emf of the cell depends on the difference in concentrations of ${M}^{2+}$ ions at the two electrodes.
The emf of the cell at $298K$ is $0.099V$
The solubility product (${K} _{sp}:{mol}^{3}{dm}^{-9}$) of ${MX} _{2}$ at $298K$ based on the information available for the given concentration cell is: (take $2.303\times R\times 298/F=0.059V$)

electrical cells patterns and properties of metals chemistry

  1. $1\times {10}^{-15}$

  2. $4\times {10}^{-15}$

  3. $1\times {10}^{-12}$

  4. $4\times {10}^{-12}$

Show answer
Correct Option: A

One part of an element (A) combines with two parts of another element (B). Six parts of element (C) combines with 4 parts of (B). If (A) and (C) combine together, then the ratio of their weights will be governed by:

law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

  1. law of definite proportion

  2. law of multiple proportion

  3. law of reciprocal proportion

  4. law of conservation of mass

Show answer
Correct Option: C
Explanation:

Law of reciprocal proportion states that when two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so will be the same or some simple multiple of the mass ratio in which they combine with each other.

Different proportions of oxygen in the various oxides of nitrogen, prove the law of:

law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

  1. reciprocal proportions

  2. multiple proportions

  3. constant proportions

  4. conservation of mass

Show answer
Correct Option: B
Explanation:

Different proportion of oxygen in the various oxides of nitrogen proves the law of multiple proportions, which states:

when two elements combine in more than one proportion to form one or more compounds, the weight of one element that combine with the given weight of other elements are in the ratio of small whole number.


Which one of the following sets of compound correctly illustrates the law of reciprocal proportions?

law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

  1. $P _2O _3, PH _3, H _2O$

  2. $P _2O _5, PH _3, H _2O$

  3. $N _2O _5, NH _3, H _2O$

  4. $N _2O, NH _3, H _2O$

  5. $NO _2, NH _3, H _2O$

Show answer
Correct Option: B
Explanation:

In $PH _3$, the ratio by weight of $P:H=31:3$

For $H _2O, O:H= 16:2=8:1$
Keeping the weight of $H(=1)$ fixed, $P:O=\cfrac {31}{3}:\cfrac {8}{1}=31:24\longrightarrow (1)$
In $P _2O _5, P:O$ is $(2 \times 31):(5 \times 16)$
$=62:80$ or $31:40 \longrightarrow (2)$
Keeping the weight of $P(=31)$ fixed in equation (1) & (2), the ratio of oxygen is $24:40$ or $3:5$ which is a simple ratio.

One of the following combinations illustrates the law of reciprocal proportion-

law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

  1. ${N _2}{O _3},{N _2}{O _4}{N _2}{O _5}$

  2. $PaCI,NaBr,NaI$

  3. $N{S _2},C{O _2},S{O _2}$

  4. $P{H _3},{P _2}{O _3},{P _2}{O _5}$

Show answer
Correct Option: A

$H _2S$ contains $94.11\%$ sulphur; $SO _2$ contains $50\%$ oxygen and $H _2O$ contains $11.11\%$ hydrogen. Thus :

law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

  1. law of multiple proportions is followed.

  2. law of reciprocal proportions is followed.

  3. law of conservation of mass is followed.

  4. all of the above are followed

Show answer
Correct Option: B
Explanation:

The Law of Multiple Proportions : It was given by Dalton. When one element combines with the other element to form two or more different compounds, the mass of one elements, which combines with a constant mass of the other, bear a simple ratio to one another.
$N _2O, NO, NO _2$
The law of reciprocal proportions states that, when two different elements separately combine with a fixed mass of a third element, the proportions in which they combine with one another shall be either in the same ratio or some multiple of it. For example, sulphur and carbon both form compounds with hydrogen. In methane 12g of carbon react with 4g of hydrogen. In hydrogen sulphide, 32g of sulphur react with 2g of hydrogen (i.e. 64g of S for 4g of hydrogen). Sulphur and carbon form a compound in which the C: S ratio is 12:64 (i.e. $CS _2$).