Electrochemistry(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

ELECTROCHEMISTRY MCQs

Total Questions : 30 | Page 1 of 3 pages

Given standard electrode potentials
$F e^{+ +} + 2 e^{-} \to F e; E^{\circ} = - 0.440 V$
$F e^{+ + +} + 3 e^{-} \to F e; E^{\circ} = - 0.036 V$
The standard electrode potential $(E^{\circ})$ for $F e^{+ + +} + e^{-} \to F e^{+ +}$ is

- 0.476 V
- 0.404 V
+ 0.404 V
+ 0.772 V

Discuss Question

Answer: Option D. -> + 0.772 V
:
D

$Δ G^{\circ} = - n E^{\circ} F$
$F e^{2 +} + 2 e^{-} \to F e$ $. . . . . . . (i)$
$Δ G^{\circ} = - 2 \times F \times (- 0.440 V) = 0.880 F$
$F e^{3 +} + 3 e^{-} \to F e$ $. . . . . . (i i)$
$Δ G^{\circ} = - 3 \times F \times (- 0.036) = 0.108 F$
On substracting equation $(i)$ from $(i i)$
$F e^{3 +} + e^{-} \to F e^{2 +}$
$Δ G^{\circ} = 0.108 F - 0.880 F = - 0.772 F$
$E^{\circ}$ for the reaction $= - \frac{Δ G^{\circ}}{n F} = - \frac{(- 0.772 F)}{1 \times F} = + 0.772 V$

Question 2.

Calculate standard free energy change for the reaction $\frac{1}{2} C u (s) + \frac{1}{2} C l_{2} (g) ⇌ \frac{1}{2} C u^{2 +} + C l^{-}$ taking place at in a cell whose standard e.m.f. is $1.02 v o l t s$

- 98430 J
98430 J
96500 J
- 49215 J

Discuss Question

Answer: Option A. -> - 98430 J
:
A

$Δ G = - n F E^{\circ}$
$\Rightarrow Δ G = - 1 \times 96500 \times 1.02 \Rightarrow Δ G = - 98430$

Question 3.

$E^{\circ}$ for the cell is $Z n | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u$ at $25^{\circ} C$ , the equilibrium constant for the reaction $Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)$ is of the order of

$10^{- 37}$
$10^{- 28}$
$10^{+ 18}$
$10^{+ 17}$

Discuss Question

Answer: Option A. ->

10^{- 37}

:
A

$E_{c e l l}^{\circ} = \frac{0.059}{n} l o g K$
$l o g K = \frac{1.10 \times 2}{0.059} = 37.2881 \Rightarrow K = 10^{- 37}$

Question 4.

In a cell that utilizes the reaction $Z n_{(s)} + 2 H^{+} (a q) \to Z n^{2 +} (a q) + H_{2 (g)}$ addition of $H_{2} S O_{4}$ to cathode compartment, will

Increase the E and shift equilibrium to the right
Lower the E and shift equilibrium to the right
Lower the E and shift equilibrium to the left
Increase the E and shift equilibrium to the left

Discuss Question

Answer: Option A. -> Increase the E and shift equilibrium to the right
:
A

$Z n_{(s)} + 2 H_{(a q)}^{+} ⇌ Z n_{(a q)}^{2 +} + H_{2 (g)}$
$E_{c e l l} = E_{c e l l}^{\circ} - \frac{.059}{2} l o g \frac{[Z n^{2 +}]}{[H^{+}]^{2}}$
When $H_{2} S O_{4}$ is added then $[H^{+}]$ will increase therefore $E_{c e l l}$ will also increases and equilibrium will shift towards right.

Question 5.

$K M n O_{4}$ acts as an oxidising agent in the neutral medium and gets reduced to $M n O_{2}$ .The equivalent weight of $K M n O_{4}$ in neutral medium

mol. wt/2
mol. wt/3
mol. wt/4
mol .wt/7

Discuss Question

Answer: Option B. -> mol. wt/3
:
B

In neutral medium $M n^{+ 7}$ oxidation state changes into +4 oxidation state, hence equivalent weight of $K M n O_{4} = \frac{M}{3}$

Question 6.

In the experiment set up for the measurement of EMF of a half cell using a reference electrode and a salt bridge, when the salt bridge is removed, the voltage

Does not change
Decreases to half the value
Increase to maximum
Drops to zero

Discuss Question

Answer: Option D. -> Drops to zero
:
D

The salt bridge is required to maintain electrical neutrality. When the salt bridge is removed, charges would accumulate at the electrodes which prevents production of electricity.

Question 7.

The rusting of iron takes place as follows $2 H^{+} + 2 e^{-} + \frac{1}{2} O_{2} \to H_{2} O (l); E^{\circ} = + 1.23 V$ $F e^{2 +} + 2 e^{-} \to F e (s); E^{\circ} = - 0.44 V$ Calculate $Δ G^{\circ}$ for the net process

$- 322 k J m o l^{- 1}$
$- 161 k J m o l^{- 1}$
$- 152 k J m o l^{- 1}$
$76 k J m o l^{- 1}$

Discuss Question

Answer: Option A. ->

- 322 k J m o l^{- 1}

:
A

\begin{matrix} F e (s) & ⟶ & F e^{2 +} & + & 2 e^{-}; & Δ G_{1}^{\circ} 2 H^{+} & + & 2 e^{-} & + & \frac{1}{2} O_{2} & ⟶ & H_{2} O (I); & Δ G_{2}^{\circ} F e (s) & + & 2 H^{+} & + & \frac{1}{2} O_{2} & ⟶ & F e^{2 +} & + & H_{2} O & ; Δ G_{3}^{\circ} \end{matrix}

Applying,

Δ G_{1}^{\circ} + Δ G_{2}^{\circ} = Δ G_{3}^{\circ}

Δ G_{3}^{\circ} = (- 2 F \times 0.44) + (- 2 F \times 1.23)

Δ G_{3}^{\circ} = - (2 \times 96500 \times 0.44 + 2 \times 96500 \times 1.23)

Δ G_{3}^{\circ} = - 322310 J

∴ Δ G_{3}^{\circ} = - 322 k J

Question 8.

The limiting molar conductivities $λ^{\circ}$ for $N a C l, K B r$ and $K C l$ are $126, 152$ and $150 S c m^{2} m o l^{- 1}$ respectively. The $λ^{\circ}$ for $N a B r$ is

$278 S c m^{2} m o l^{- 1}$
$176 S c m^{2} m o l^{- 1}$
$128 S c m^{2} m o l^{- 1}$
$302 S c m^{2} m o l^{-} 1$

Discuss Question

Answer: Option C. ->

128 S c m^{2} m o l^{- 1}

:
C

$(126 s c m^{2}) \land_{N a C l}^{\circ} = \land_{N a +}^{\circ} + \land_{C l^{-}}^{\circ}$                  $. . . . . . . (1)$
$(152 s c m^{2}) \land_{K B r}^{\circ} = \land_{K^{+}}^{\circ} + \land_{B r^{-}}^{\circ}$                        $. . . . . . . (2)$
$(150 s c m^{2}) \land_{K C l}^{\circ} = \land_{K^{+}}^{\circ} + \land_{C l^{-}}^{\circ}$                           $. . . . . . . (3)$
By equation $(1) + (2) - (3)$
$∵ \land_{N a B r}^{\circ} = \land_{N a +}^{\circ} + \land_{B r^{-}}^{\circ} = 126 + 152 - 150 = 128 S c m^{2} m o l^{- 1}$

Question 9.

$4.5 g$ of aluminum (at. mass 27amu) is deposited at cathode from $A l^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H +$ ions in solution by the same quantity of electric charge will be

22.4 L
44.8 L
5.6 L
11.2 L

Discuss Question

Answer: Option C. -> 5.6 L
:
C

E q o f A l = e q o f H_{2}

\frac{4.5}{\frac{27}{3}} = e q o f H_{2}; \frac{4.5}{9} = e q o f H_{2}

2 H^{+} + 2 e^{-} \to H_{2}

eq. of

H_{2}

= Number of moles

\times n

factor

0.5 = n_{H_{2}} \times 2

V_{H_{2}} = \frac{0.5}{2} \times 22.4; V_{H_{2}} = 5.6 L

Question 10.

Assertion: If $λ_{N a^{+}}^{\circ} + λ_{C l^{-}}^{\circ}$ are molar limiting conductivity of the sodium and chloride ions respectively, then the limiting molar conducting for sodium chloride is given by the equation: $λ_{N a C l}^{\circ} = λ_{N a^{+}}^{\circ} + λ_{C l^{-}}^{\circ}$
Reason:This is according to Kohlrausch law of independent migration of ions.

If both assertion and reason are true and the reason is the correct explanation of the assertion.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.

Discuss Question

Answer: Option A. -> If both assertion and reason are true and the reason is the correct explanation of the assertion.
:
A

According to Kohlrausch law, “Limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte”.