Electrochemistry Test

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Electrochemistry Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following options will be correct for the stage of half completion of the reaction $$A \rightleftharpoons B$$?

A
$$\triangle G^0 = 0$$

B
$$\triangle G^0 > 0$$

C
$$\triangle G^0 < 0$$

D
$$\triangle G^0 = -RT ln 2$$

Solution

When the reaction is half completed, [A] = [B].
Thus, $$\frac{[B]}{[A]} = 1$$
$$\triangle G^0 = -RTInK = -RTIn(1) = 0$$
Question 2
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$${ E }^{ 0 }$$ value of $$Mg/{ Mg }^{ +2 }$$ is $$+2.37$$ V of $$Zn/{ Zn }^{ +2 }$$ is $$+0.76$$ V and that of $$Fe/{ Fe}^{ +2 }$$ is $$+0.44$$ V, which of the following statements is correct?

A
Zinc will reduce $${ Mg }^{ +2 }$$

B
Zn will reduce $${ Fe }^{ +2 }$$

C
Mg oxidizes Fe

D
Zn oxidizes Fe

Solution
Question 3
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Given the standard half-cell potentials $$(E^o)$$ of the following as

$$Zn=Zn^{2+}+2e^-$$; $$E^o=+0.76$$V
$$Fe=Fe^{2+}+2e^{-}$$; $$E^o=+0.41$$V

Then the standard e.m.f. of the cell with the reaction $$Fe^{2+}+Zn\rightarrow Zn^{2+}+Fe$$ is?

A
$$-0.35$$V

B
$$+0.35$$V

C
$$+1.17$$V

D
$$-1.17$$V

Solution

$$Zn^{+2}+2e^-\rightarrow Zn$$ ; $$E^o=-0.76\ V$$ (SRP) (Anode)

$$Fe^{+2}+2e^-\rightarrow Fe$$ ; $$E^o=-0.41\ V$$ (SRP) (Cathode)

$$E^o_{cell}=E^+_{Fe^{+2}/Fe}-E^o_{Zn^{+2}/Zn}$$

$$=-0.41+0.76=0.35V$$.

Hence, the correct option is $$\text{B}$$
Question 4
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The quantity of charge required to obtain one mole of aluminium from $$Al_2O_3$$ is:

A
$$1$$ F

B
$$6$$ F

C
$$3$$ F

D
$$2$$ F

Solution

From $$Al_2O_3\rightarrow 2Al$$ to obtain one mole of $$Al$$ we consider,
$$2Al^{3+}+6e^{-}\rightarrow2Al$$
So, 6F of charge is required to obtain 2 moles of $$Al$$ from $$Al_2O_3$$
Thus, 3F of charge is required to obatain 1 mole of $$Al$$ from $$Al_2O_3$$
Question 5
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What will be the reduction potential for the following half-cell reaction at $$298$$K?
(Given: $$[Ag^+]=0.1$$M and $$E^o_{cell}=+0.80$$V)

A
$$0.741$$V

B
$$0.80$$V

C
$$-0.80$$V

D
$$-0.741$$V

Solution

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

Nernst equation is,

$$E_{cell}=E^{0}_{cell}-\dfrac{0.0591}{n}logQ$$

Here the reaction is,

$$Ag^{+}+e^{-}\rightarrow Ag$$

$$n=1$$

$$E^{0}_{cell}=0.80V$$

Hence,

$$E_{cell}=E^{0}_{cell}-\dfrac{0.0591}{n}log\dfrac{1}{[Ag^{+}]}$$

$$E_{cell}=0.80-\dfrac{0.0591}{1}log\dfrac{1}{0.1}$$

$$=0.741V$$
Question 6
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$$\Delta G^o$$ for the reaction, $$Cu^{2+}+Fe\rightarrow Fe^{2+}+Cu$$ is:

(Given: $$E^o_{Cu^{2+}|Cu}=+0.34$$V, $$E^o_{Fe^{2+}|Fe}=-0.44$$V).

A
$$11.44\ J$$

B
$$180.8\ kJ$$

C
$$150.5\ kJ$$

D
$$28.5\ kJ$$

Solution

We know that,

$$ E^{0}_{cell}=E^{0}_{Cu^{2+}/Cu}-E^{0}_{Fe^{2+}/Fe}$$

$$=0.34-(-0.44)V=0.78V$$

Now,

$$\Delta G=-nFE^{0}_{cell}$$

$$=-2\times96500\times0.78=150540\ J$$ or $$ 150.5\ kJ$$
Question 7
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At $$1000^0C$$,
$$Zn_(s)+\frac{1}{2}\rightarrow ZnO_{(s)}; \triangle G^0 = - 360 kJ mol^{-1}$$
$$C_{(1)} +\frac{1}{2} O_{2(g)}; \triangle G^0=-460kJmol^{-1}$$
The correct statement is:

A
zinc can be oxidised by carbon monoxide

B
zinc blend is produced during the reaction

C
zinc oxide can be reduced by graphite

D
zinc can be oxidised by graphite

Solution

Formation of $$'CO'$$, is more spontaneous as it has more negative $$\triangle G$$ value. Then $$ZnO$$ changes to $$Zn$$ because of less negative $$'\triangle G'$$ value. So, $$ZnO$$ can be reduced by Graphite $$ZnO+C\rightarrow Zn+CO$$
Question 8
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Which of the following is the cell reaction that occurs when the following half-cells are combined?
$$I_2+2e^-\rightarrow 2I^-(1M)$$; $$E^o=+0.54$$V
$$Br_2+2e^-\rightarrow 2Br^-(1M); E^o=+1.09$$V

A
$$2Br^-+I_2\rightarrow Br_2+2I^-$$

B
$$I_2+Br_2\rightarrow 2I^-+2Br^-$$

C
$$2I^-+Br_2\rightarrow I_2+2Br^-$$

D
$$2I^-+2Br^-\rightarrow I_2Br_2$$

Solution

A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.
Reduction potential of both $$I_2$$ and $$Br_2$$ is given. We have to note that more the value of reduction potential then less the element has the tendency to get reduced and hence $$I_2$$ having more reduction potential acts as anode i.e. undergoes oxidation. And $$Br_2$$ having less reduction potential than $$I_2$$ undergoes reduction and acts as cathode.

$$2I^-\rightarrow I_2+2e^-$$ (Oxidation)
$$Br_2+2e^{-}\rightarrow 2Br^-$$ (Reduction)
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$
$$2I^-+Br_2\rightarrow I-2+2Br^-$$ is net cell reaction
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$

Hence option (C) is correct.
Question 9
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The specific conductance of a saturated solution of silver bromide is $$\kappa \,{\text{S}}\,{\text{c}}{{\text{m}}^{ - 1}}$$. The limiting ionic conductivity of $${\text{A}}{{\text{g}}^ + }\,{\text{and}}\,{\text{B}}{{\text{r}}^ - }$$ ions are x and y, respectively. The solubility of silver bromide in $${\text{g}}{{\text{L}}^{{\text{ - 1}}}}$$ is: [molar mass of AgBr = 188]

A
$$\dfrac{{\kappa \times 1000}}{{{\text{x}} - {\text{y}}}}$$

B
$$\dfrac{\kappa }{{{\text{x}} + {\text{y}}}} \times 188$$

C
$$\dfrac{{\kappa \times 1000 \times 188}}{{{\text{x}} + {\text{y}}}}$$

D
$$\dfrac{{{\text{x}} \times {\text{y}}}}{\kappa } \times \dfrac{{1000}}{{188}}$$

Solution

Given Specific conductance of AgBr = K S $${\text{c}}{{\text{m}}^{ - 1}}$$

$$\Lambda {^\circ _{A{g^ + }}} = x$$   $$\Lambda {^\circ _{B{r^ - }}} = y$$

  $$\Lambda {^\circ _{AgBr}} = \Lambda {^\circ _{A{g^ + }}} + \Lambda {^\circ _{B{r^ - }}} = x + y$$

$$\Lambda {^\circ _{AgBr}} = {\text{K}} \times \dfrac{{1000}}{{{\text{C}}\left( {{\text{mol}}\,{{\text{L}}^{ - 1}}} \right)}}$$

  $$x + y = \dfrac{{{\text{K}} \times 1000}}{{{\text{C}}\left( {{\text{mol}}\,{{\text{L}}^{ - 1}}} \right)}}$$

(Solubility) C = $$\left( {\dfrac{{K \times 1000}}{{x + y}}} \right){\text{mol}}\,.{{\text{L}}^{ - 1}}$$

No. of mol = $$\dfrac{{{\text{mass(g)}}}}{{{\text{molar}}\,\,{\text{mass}}\left( {{\text{g}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)}}$$

molar mass of AgBr = 188 $${{\text{g}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}$$
So, $$\boxed{{\text{C}} = \left( {\dfrac{{{\text{K}} \times 1000}}{{x + y}}} \right) \times 188\,\,{\text{g}}{\text{.}}{{\text{L}}^{ - 1}}}$$

Hence, option (C) is correct.
Question 10
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$$E^o$$ values for the half cell reactions are given:
$$Cu^{2+}+e^-\rightarrow Cu^+$$; $$E^o=0.15$$V
$$Cu^{2+}+2e^-\rightarrow Cu$$; $$E^o=0.34$$V
What will be the $$E^o$$ of the half-cell: $$Cu^{+}+e^-\rightarrow Cu$$?

A
$$+0.49$$V

B
$$+0.19$$V

C
$$+0.53$$V

D
$$+0.30$$V

Solution

Given,
$$Cu^{2+}+e^{-}\rightarrow Cu^{+}; E^{0}_{1}=0.15V,\Delta G^{0}_{1},n_1=1$$

$$Cu^{2+}+2e^{-}\rightarrow Cu; E^{0}_{2}=0.34V,\Delta G^{0}_{2},n_2=2$$

$$Cu^{+}+e^{-}\rightarrow Cu; E^{0}_{3}=?V,\Delta G^{0}_{3},n_3=1$$

As $$\Delta G$$ is an extensive property,

$$\Delta G^{0}_{3}=\Delta G^{0}_{2}-\Delta G^{0}_{1}$$

$$\implies$$ $$ -n_3FE^{0}_{3}=-n_2FE^{0}_{2}+ n_1FE^{0}_{1}$$

$$-E^{0}_{3}=-2\times0.34+1\times0.15$$

$$E^{0}_{3}=0.68-0.15=0.53V$$

Hence, option C is correct.

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Electrochemistry Test - 62

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Electrochemistry Test - 62

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Question 1

$$\triangle G^0 = 0$$

$$\triangle G^0 > 0$$

$$\triangle G^0 < 0$$

$$\triangle G^0 = -RT ln 2$$

Solution

Question 2

Zinc will reduce $${ Mg }^{ +2 }$$

Zn will reduce $${ Fe }^{ +2 }$$

Mg oxidizes Fe

Zn oxidizes Fe

Solution

Question 3

$$-0.35$$V

$$+0.35$$V

$$+1.17$$V

$$-1.17$$V

Solution

Question 4

$$1$$ F

$$6$$ F

$$3$$ F

$$2$$ F

Solution

Question 5

$$0.741$$V

$$0.80$$V

$$-0.80$$V

$$-0.741$$V

Solution

Question 6

$$11.44\ J$$

$$180.8\ kJ$$

$$150.5\ kJ$$

$$28.5\ kJ$$

Solution

Question 7

zinc can be oxidised by carbon monoxide

zinc blend is produced during the reaction

zinc oxide can be reduced by graphite

zinc can be oxidised by graphite

Solution

Question 8

$$2Br^-+I_2\rightarrow Br_2+2I^-$$

$$I_2+Br_2\rightarrow 2I^-+2Br^-$$

$$2I^-+Br_2\rightarrow I_2+2Br^-$$

$$2I^-+2Br^-\rightarrow I_2Br_2$$

Solution

Question 9

$$\dfrac{{\kappa \times 1000}}{{{\text{x}} - {\text{y}}}}$$

$$\dfrac{\kappa }{{{\text{x}} + {\text{y}}}} \times 188$$

$$\dfrac{{\kappa \times 1000 \times 188}}{{{\text{x}} + {\text{y}}}}$$

$$\dfrac{{{\text{x}} \times {\text{y}}}}{\kappa } \times \dfrac{{1000}}{{188}}$$

Solution

Question 10

$$+0.49$$V

$$+0.19$$V

$$+0.53$$V

$$+0.30$$V

Solution

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