22. ⇒ (JEE Main 2022 (Online) 29th June Evening Shift )

The cell potential for the given cell at 298 K

Pt| H₂ (g, 1 bar) | H⁺ (aq) || Cu²⁺ (aq) | Cu(s)

is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10^{$-$ x} M. The value of x is ___________.

(Given : $E_{C u^{2 +} / C u}^{Θ}$ = 0.34 V and $\frac{2.303 R T}{F}$ = 0.06 V)

correct answer is 7

# Explanation

$Q = \frac{{[H^{+}]}^{2}}{[{Cu}^{+ 2}] {pH}_{2}} = \frac{10^{- 6}}{C} {pH}_{2} = 1$

$\begin{aligned} E = E_{cell}^{\circ} - \frac{0.06}{n} \log Q \\ 0.31 = 0.34 - \frac{0.06}{2} \log \frac{10^{- 6}}{C} \\ \log \frac{10^{- 6}}{C} = 1 \\ C = 10^{- 7} M \\ x = 7 \end{aligned}$

23. ⇒ (JEE Main 2022 (Online) 28th June Evening Shift )

For the given reactions

Sn²⁺ + 2e^$-$ $\to$ Sn

Sn⁴⁺ + 4e^$-$ $\to$ Sn

the electrode potentials are ; $E_{S n^{2 +} / S n}^{o} = - 0.140$ V and $E_{S n^{4 +} / S n}^{o} = + 0.010$ V. The magnitude of standard electrode potential for $S n^{4 +} / S n^{2 +}$ i.e. $E_{S n^{4 +} / S n^{2 +}}^{o}$ is _____________ $\times$ 10^{$-$ 2} V. (Nearest integer)

correct answer is 16

# Explanation

$Sn ⟶ {Sn}^{2 +} + 2 e^{-} E_{1}^{0} = 0.140 V$

${Sn}^{4 +} + 4 e^{-} ⟶ Sn E_{2}^{0} = 0.010 V$

$\begin{aligned} {Sn}^{4 +} + 2 e^{-} ⟶ {Sn}^{2 +} E_{cell}^{0} \\ E_{cell}^{O} = \frac{n_{2} E_{2}^{o} + n_{1} E_{1}^{0}}{n} = \frac{4 (0.010) + 2 (0.140)}{2} \\ E_{cell}^{0} = 0.16 V = 16 \times 10^{- 2} V \end{aligned}$

24. ⇒ (JEE Main 2022 (Online) 27th June Evening Shift )

For the reaction taking place in the cell :

Pt (s)| H₂ (g)|H⁺(aq) || Ag⁺(aq) |Ag (s)

E $_{c e l l}^{o}$ = + 0.5332 V.

The value of $Δ$ _fG $^{\circ}$ is ______________ kJ mol^{$-$ 1}. (in nearest integer)

correct answer is 51 or 103

# Explanation

JEE Main 2022 (Online) 27th June Evening Shift Chemistry - Electrochemistry Question 46 English Explanation

# At anode, oxidation occur

H₂ $\to$ 2H⁺ + 2e^$-$ ....... (1)

# At cathode, reduction occur

2Ag⁺ + 2e^$-$ $\to$ 2Ag ...... (2)

Adding equation (1) and (2), we get n = 2, where n = cancelled out electron

Now,

$Δ G^{\circ} = - n F E_{c e l l}^{o}$

$= - 2 \times 96500 \times 0.5332$

$= - 102907.6$

$= - 102.9$ kJ/mol

$= - 103$ kJ/mol

25. ⇒ (JEE Main 2022 (Online) 26th June Morning Shift )

The ${(\frac{\partial E}{\partial T})}_{P}$ of different types of half cells are as follows:

A	B	C	D
$1 \times 10^{- 4}$	$2 \times 10^{- 4}$	$0.1 \times 10^{- 4}$	$0.2 \times 10^{- 4}$

(Where E is the electromotive force)

Which of the above half cells would be preferred to be used as reference electrode?

(A) A

(B) B

(C) C

(D) D

Explanation

Correct answer is C

A cell with less variation in EMF with temperature is preferred as a reference electrode because it can be used for a wider range of temperatures without much derivation from standard value so a cell with less ${(\frac{\partial E}{\partial T})}_{P}$ is preferred.

26. ⇒ (JEE Main 2022 (Online) 26th June Evening Shift )

Cu(s) + Sn²⁺ (0.001M) $\to$ Cu²⁺ (0.01M) + Sn(s)

The Gibbs free energy change for the above reaction at 298 K is x $\times$ 10^{$-$ 1} kJ mol^{$-$ 1}. The value of x is __________. [nearest integer]

[Given : $E_{C u^{2 +} / C u}^{Θ} = 0.34 V$ ; $E_{S n^{2 +} / S n}^{Θ} = - 0.14 V$ ; F = 96500 C mol^{$-$ 1}]

correct answer is 983

# Explanation

$\begin{aligned} Cu & + {Sn}^{+ 2} ⟶ {Cu}^{+ 2} + Sn (s) \\ E_{cell}^{\circ} & = E_{OX}^{\circ} + E_{Red}^{\circ} \\ = - 0.34 - 0.14 \\ = - 0.48 V \\ E = & E^{\circ} - \frac{0.0591}{2} \log \frac{[{Cu}^{+ 2}]}{[{Sn}^{+ 2}]} \\ = & - 0.48 - 0.0295 \log 10 \\ = & - 0.5095 V \\ Δ G = & - nFE \\ = - 2 \times 96500 \times - 0.5095 J / mol \\ = 98333.5 \times 10^{- 3} kJ / mol \\ = 983.3 \times 10^{- 1} kJ / mol \\ = 983 \times 10^{- 1} kJ / mol \end{aligned}$

27. ⇒ (JEE Main 2022 (Online) 25th June Morning Shift )

In a cell, the following reactions take place

$\begin{matrix} F e^{2 +} \to F e^{3 +} + e^{-} & E_{F e^{3 +} / F e^{2 +}}^{o} = 0.77 V \\ 2 I^{-} \to I_{2} + 2 e^{-} & E_{I_{2} / I^{-}}^{o} = 0.54 V \end{matrix}$

The standard electrode potential for the spontaneous reaction in the cell is x $\times$ 10^{$-$ 2} V 298 K. The value of x is ____________. (Nearest Integer)

correct answer is 23

# Explanation

$\underset{Cathode}{{Fe}^{+ 3}} + \underset{anode}{I^{-}} ⟶ I_{2} + {Fe}^{+ 2}$

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

$\begin{aligned} = 0.77 - 0.54 \\ = 0.23 V \\ = 23 \times 10^{- 2} V \end{aligned}$

28. ⇒ (JEE Main 2022 (Online) 25th June Evening Shift )

The correct order of reduction potentials of the following pairs is

A. Cl₂/Cl^$-$

B. I²/I^$-$

C. Ag⁺/Ag

D. Na⁺/Na

E. Li⁺/Li

Choose the correct answer from the options given below.

(A) A > C > B > D > E

(B) A > B > C > D > E

(C) A > C > B > E > D

(D) A > B > C > E > D

Explanation

Correct answer is A

$E_{C_{2} / CI}^{\circ} = + 1.36 V$

$E_{I_{2} / I^{-}}^{0} = + 0.54 V$

$E_{{Ag}^{+} / Ag}^{\circ} = + 0.80 V$

$E_{{Na}^{+} / Na}^{\circ} = - 2.71 V$

$E_{L^{+} / Li} = - 3.05 V$

29. ⇒ (JEE Main 2022 (Online) 24th June Morning Shift )

The cell potential for the following cell

Pt |H₂(g)|H⁺ (aq)|| Cu²⁺ (0.01 M)|Cu(s)

is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)

(Given : $E_{C u^{2 +} / C u}^{o} = 0.34$ V and $\frac{2.303 R T}{F} = 0.06$ V)

correct answer is 5

# Explanation

$E_{cell} = E_{cell}^{0} - \frac{0.06}{2} \log \frac{{[H^{\oplus}]}^{2}}{[{Cu}^{+ 2}]}$

$0.576 = 0.34 - 0.03 \log \frac{{[H^{\oplus}]}^{2}}{[0.01]}$

$0.576 - 0.34 = - 0.03 \log {[H^{\oplus}]}^{2} + 0.03 \log (0.01)$

$= 0.06 pH - 0.06$

$pH ≃ 4.93 ≃ 5$

30. ⇒ (JEE Main 2021 (Online) 31st August Morning Shift )

Consider the following cell reaction :

$C d_{(s)} + H g_{2} S O_{4 (s)} + \frac{9}{5} H_{2} O_{(l)}$ $⇌$ $C d S O_{4} . \frac{9}{5} H_{2} O_{(s)} + 2 H g_{(l)}$

The value of $E_{c e l l}^{0}$ is 4.315 V at 25 $^{\circ}$ C. If $Δ$ H $^{\circ}$ = $-$ 825.2 kJ mol^{$-$ 1}, the standard entropy change $Δ$ S $^{\circ}$ in J K^{$-$ 1} is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol^{$-$ 1}]

correct answer is 25

# Explanation

$Δ G^{\circ} = - n F E^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$

$∴$ $Δ$ S $^{\circ}$ $= \frac{Δ H^{\circ} + n F E^{\circ}}{T}$

$= \frac{(- 825.2 \times 10^{3}) + (2 \times 96487 \times 4.315)}{298}$

$= \frac{- 825.2 \times 10^{3} + 832.682 \times 10^{3}}{298}$

$= \frac{7.483 \times 10^{3}}{298} = 25.11$ JK^{$-$ 1} mol^{$-$ 1}

$∴$ Nearest integer answer is 25.

31. ⇒ (JEE Main 2021 (Online) 26th August Evening Shift )

For the galvanic cell,

Zn(s) + Cu²⁺ (0.02 M) $\to$ Zn²⁺ (0.04 M) + Cu(s),

E_cell = ______________ $\times$ 10^{$-$ 2} V. (Nearest integer)

[Use : $E_{C u / C u^{2 +}}^{0}$ = $-$ 0.34 V, $E_{Z n / Z n^{2 +}}^{0}$ = + 0.76 V, $\frac{2.303 R T}{F} = 0.059 V$ ]

correct answer is 109

# Explanation

Galvanic cell :

$Z n_{(s)} + \underset{0.02 M}{C u_{(a q .)}^{+ 2}} \to \underset{0.04 M}{Z n_{}^{+ 2}} + C u (s)$

Nernst equation = $F_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{2} \log \frac{[2 n^{+ 2}]}{[C u^{+ 2}]}$

$\Rightarrow E_{c e l l} [E_{c e l l}^{o} - E_{Z n^{+ 2} / Z n}^{o}] - \frac{0.059}{2} \log \frac{0.04}{0.02}$

$\Rightarrow E_{c e l l} [0.34 - (- 0.76)] - \frac{0.059}{2} \log^{2}$

$\Rightarrow E_{c e l l} 1 - 1 - \frac{0.059}{2} \times 0.3010$

= 1.0911 = 109.11 $\times$ 10^{$-$ 2}

= 109

32. ⇒ (JEE Main 2021 (Online) 26th August Morning Shift )

These are physical properties of an element

(A) Sublimation enthalpy

(B) Ionisation enthalpy

(C) Hydration enthalpy

(D) Electron gain enthalpy

The total number of above properties that affect the reduction potential is ____________ (Integer answer)

correct answer is 3

# Explanation

Sublimation enthalpy, Ionisation enthalpy and hydration enthalpy affect the reduction potential.

⇒Electrochemistry

Topic 3: Cells and Electrode Potential, Nernst Equation 3