33. ⇒ (JEE Main 2021 (Online) 27th July Evening Shift )

For the cell

Cu(s) | Cu²⁺ (aq) (0.1 M) || Ag⁺(aq) (0.01 M) | Ag(s)

the cell potential E₁ = 0.3095 V

For the cell

Cu(s) | Cu²⁺ (aq) (0.01 M) || Ag⁺(aq) (0.001 M) | Ag(s)

the cell potential = ____________ $\times$ 10^{$-$ 2} V. (Round off the nearest integer).

[Use : $\frac{2.303 R T}{F}$ = 0.059]

correct answer is 28

# Explanation

Cell reaction is :

$C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)$

Now, $E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{2} \log \frac{[C u^{2 +}]}{{[A g^{+}]}^{2}}$ .... (1)

$∴$ $E_{1} = 0.3095 = E_{c e l l}^{o} - \frac{0.059}{2} . \log \frac{0.01}{{(0.001)}^{2}}$ ....(2)

From (1) and (2), E₂ = 0.28 V = 28 $\times$ 10^{$-$ 2} V

34. ⇒ (JEE Main 2021 (Online) 25th July Morning Shift )

Consider the cell at 25 $^{\circ}$ C

Zn | Zn²⁺ (aq), (1M) || Fe³⁺ (aq), Fe²⁺ (aq) | Pt(s)

The fraction of total iron present as Fe³⁺ ion at the cell potential of 1.500 V is x $\times$ 10^{$-$ 2}. The value of x is ______________. (Nearest integer)

(Given : $E_{F e^{3 +} / F e^{2 +}}^{0} = 0.77 V$ , $E_{Z n^{2 +} / Z n}^{0} = - 0.76 V$ )

correct answer is 24

# Explanation

$Z n \underset{}{⟶} Z n^{2 +} + 2 e^{-}$

$2 F e^{3 +} \underset{}{⟶} 2 e^{-} + 2 e^{2 +}$

$Z n + 2 F e^{3 +} \overset{}{⟶} Z n^{2 +} + 2 F e^{2 +}$

$E_{c e l l}^{0} = 0.77 - (0.76)$

$= 1.53$ V

$1.50 = 1.53 - \frac{0.06}{2} \log {(\frac{F e^{2 +}}{F e^{3 +}})}^{2}$

$\log (\frac{F e^{2 +}}{F e^{3 +}}) = \frac{0.03}{0.06} = \frac{1}{2}$

$\frac{[F e^{2 +}]}{[F e^{3 +}]} = 10^{1 / 2} = \sqrt{10}$

$\frac{[F e^{3 +}]}{[F e^{2 +}]} = \frac{1}{\sqrt{10}}$

$\frac{[F e^{3 +}]}{[F e^{2 +}] + [F e^{3 +}]} = \frac{1}{1 + \sqrt{10}} = \frac{1}{4.16}$

= 0.2402

= 24 $\times$ 10^-2

35. ⇒ (JEE Main 2021 (Online) 22th July Evening Shift )

Assume a cell with the following reaction

$C u_{(s)} + 2 A g^{+} (1 \times 10^{- 3} M) \to C u^{2 +} (0.250 M) + 2 A g_{(s)}$

$E_{c e l l}^{Θ} = 2.97$ V

E_cell for the above reaction is ______________ V. (Nearest integer)

[Given : log 2.5 = 0.3979, T = 298 K]

correct answer is 3

# Explanation

$E = E^{o} - \frac{0.059}{2} \log \frac{[C u^{+ 2}]}{{[A g^{+}]}^{2}}$

$= 2.97 - \frac{0.059}{2} \log \frac{0.25}{{(10^{- 3})}^{2}} = 2.81 V$

36. ⇒ (JEE Main 2021 (Online) 18th March Morning Shift )

For the reaction

2Fe³⁺(aq) + 2I^$-$(aq) $\to$ 2Fe²⁺(aq) + I₂(s)

the magnitude of the standard molar Gibbs free energy change, $Δ$ _rG $_{m}^{o}$ = $-$ ___________ kJ (Round off to the Nearest Integer).

$[\begin{matrix} E_{F e^{2 +} / F e (s)}^{o} = - 0.440 V; & E_{F e^{3 +} / F e (s)}^{o} = - 0.036 V \\ E_{I_{2} / 2 I^{-}}^{o} = 0.539 V; & F = 96500 C \end{matrix}]$

correct answer is 46

# Explanation

$E_{F e^{3 +} / F e}^{o} = 0.036 V$

$F e^{3 +} + 3 e^{-} \to F e$

$∴$ $Δ G_{1}^{o} = - n F E^{o}$

$= - 3 F (- 0.036)$

$E_{F e^{2 +} / F e}^{o} = 0.440 V$

$F e^{2 +} + 2 e^{-} \to F e; E^{o} = - 0.440 V$

$F e \to F e^{2 +} + 2 e^{-}; E^{o} = 0.440 V$

$∴$ $Δ G_{2}^{o} = - 2 F (0.440)$

$E_{I_{2} / 2 I^{-}}^{o} = 0.539 V$

$I_{2} + 2 e^{-} \to 2 I^{-}; E^{o} = 0.539 V$

$2 I^{-} \to I_{2} + 2 e^{-}; E^{o} = - 0.539 V$

$Δ G_{3}^{o} = - 2 F (- 0.539)$

$∴$ $Δ G^{o} = 2 [Δ G_{1}^{o} + Δ G_{2}^{o}] + Δ G_{3}^{o}$

$= 2 [3 F (0.036) - 2 F (0.440)] + 2 F (0.539)$

$= - 45934 = - 45.9 K J ≃ - 46 K J$

37. ⇒ (JEE Main 2021 (Online) 26th February Evening Shift )

Emf of the following cell at 298K in V is x $\times$ 10^{$-$ 2}.

Zn|Zn²⁺(0.1 M)||Ag⁺ (0.01 M)|Ag

The value of x is _________. (Rounded off to the nearest integer)

[Given : $E_{Z n^{2 +} / Z n}^{θ} = - 0.76 V; E_{A g^{2 +} / A g}^{θ} = + 0.80 V; \frac{2.303 R T}{F} = 0.059$ ]

correct answer is 147

# Explanation

Zn | Zn²⁺(0.1 M) || Ag⁺ (0.01 M) | Ag

Zn(s) + 2Ag⁺ $⇌$ 2Ag(s) + Zn⁺²

$E_{c e l l}^{0} = E_{A g^{+} / A g}^{0} - E_{Z n^{2 +} / Z n}^{0}$

$= 0.80 - (- 0.76)$

$= 1.56 V$

$E_{c e l l} = 1.56 - \frac{0.059}{2} \log \frac{[Z n^{2 +}]}{{[A g^{+}]}^{2}}$

$= 1.56 - \frac{0.059}{2} \log \frac{0.1}{{(0.01)}^{2}}$

$= 1.56 - \frac{0.059}{2} \times 3$

$= 1.56 - 0.0885$

$= 1.4715$

$= 147.15 \times 10^{- 2}$

38. ⇒ (JEE Main 2021 (Online) 25th February Evening Shift )

Copper reduces NO $_{3}^{-}$ into NO and NO₂ depending upon the concentration of HNO₃ in solution. (Assuming fixed [Cu²⁺] and P_NO = P_NO_₂), the HNO₃ concentration at which the thermodynamic tendency for reduction of NO $_{3}^{-}$ into NO and NO₂ by copper is same is 10^x M. The value of 2x is _______. (Rounded off to the nearest integer)

[Given, $E_{C u^{2 +} / C u}^{o} = 0.34$ V, $E_{N O_{3}^{-} / N O}^{o} = 0.96$ V, $E_{N O_{3}^{-} / N O_{2}}^{o} = 0.79$ V and at 298 K, $\frac{R T}{F}$ (2.303) = 0.059]

correct answer is 1

# Explanation

Cell-I $(H N O_{3} \to N O)$

$3 C u + 2 N O_{3}^{-} + 8 H^{+} \to 3 C u^{2 +} + 2 N O + 4 H_{2} O$

$Q_{1} = \frac{{[C u^{2 +}]}^{3} \times {(p_{N O})}^{2}}{{[N O_{3}^{-}]}^{2} \times {[H^{+}]}^{8}}$

$∵$ $E_{1}^{o} = 0.96 - (- 0.34) = 1.3 V$

$E_{1} = 1.3 - \frac{0.059}{6} \log Q_{1}$

Cell-II $(H N O_{3} \to N O_{2})$

$C u + 2 N O_{3}^{-} + 4 H^{+} \to C u^{2 +} + 2 N O_{2} + 2 H_{2} O$

$Q_{2} = \frac{[C u^{2}] \times {(p_{N O_{2}})}^{2}}{{[N O_{3}^{-}]}^{2} \times {[H^{+}]}^{4}}$

$∵$ $E_{2}^{o} = 0.79 - (- 0.34) V = 1.13 V$

$E_{2} = 1.13 - \frac{0.059}{2} \log Q_{2}$

Now, $E_{1} = E_{2}$

$1.3 - \frac{0.059}{6} \log Q_{1} = 1.13 - \frac{0.059}{2} \log Q_{2}$

$0.17 = \frac{0.059}{6} [\log Q_{1} - 3 \log Q_{2} - = \frac{0.059}{6} \log \frac{Q_{1}}{Q_{2}}$

$= \frac{0.059}{6} \log \frac{{[C u^{2 +}]}^{3} \times {(p_{N O})}^{2}}{{[N O_{3}^{-}]}^{2} \times {[H^{+}]}^{8}} \times \frac{{[N O_{3}^{-}]}^{6} \times {[H^{+}]}^{12}}{{[C u^{2 +}]}^{3} \times {(p_{N O_{2}})}^{6}}$

$= \frac{0.059}{6} \log \frac{{[H^{+}]}^{4} \times {[N O_{3}^{-}]}^{4}}{{(p_{N O_{2}})}^{4}}$ [ $∵$ $p_{N O} = p_{N O_{2}}$ ]

$= \frac{0.059}{6} \log \frac{{[H N O_{3}]}^{4}}{{(p_{N O_{2}})}^{4}}$

Now, $p_{N O_{2}} \equiv [H N O_{3}]$

So, $0.17 = \frac{0.059}{6} \log [H N O_{3}]^{8}$

$= \frac{0.059}{6} \times 8 \log [H N O_{3}]$

$\log [H N O_{3}] = 2.16$

$[H N O_{3}] = 10^{2.16} M = 10^{x} M$

$∴$ $x = 2.16$

$\Rightarrow 2 x = 2 \times 2.16 = 4.32 ≃ 4$

39. ⇒ (JEE Main 2021 (Online) 24th February Evening Shift )

The magnitude of the change in oxidising power of the $M n O_{4}^{-} / M n^{2 +}$ couple is x $\times$ 10^{$-$ 4} V, if the H⁺ concentration is decreased from 1M to 10^{$-$ 4} M at 25 $^{\circ}$ C. (Assume concentration of $M n O_{4}^{-}$ and $M n^{2 +}$ to be same on change in H⁺ concentration). The value of x is ___________. $[G i v e n : \frac{2.303 R T}{F} = 0.059]$

correct answer is 3776

# Explanation

Reaction,

$M n O_{4}^{-} + H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

n = 5

Applying Nernst equation, $E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} \log \frac{[P]}{[R]}$

or $E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} \log \frac{[M n^{2 +}]}{[M n O_{4}^{-}]} {[\frac{1}{H^{+}}]}^{8}$

(I) Given, [H⁺] = 1 M

$E_{1} = E^{\circ} - \frac{0.0591}{5} \log \frac{[M n^{2 +}]}{[M n O_{4}^{-}]}$

(II) Now, [H⁺] = 10^{$-$ 4} M

$E_{2} = E^{\circ} - \frac{0.0591}{5} \log \frac{[M n^{2 +}]}{[M n O_{4}^{-}]} \times \frac{1}{{(10^{- 4})}^{8}}$

$∴$ $| E_{1} - E_{2} |$

$| E_{1} - E_{2} | = \frac{0.0591}{5} \times 32 = 0.3776 V = 3776 \times 10^{- 4}$

x = 3776

40. ⇒ (JEE Main 2021 (Online) 24th February Morning Shift )

The electrode potential of M²⁺/M of 3d-series elements shows positive value for :

(A) Zn

(B) Fe

(C) Cu

(D) Co

Explanation

Correct answer is C

In the electrode potential series, only copper have positive value for electrode potential because copper has lower tendency than hydrogen to form ions. So, if standard hydrogen electrode (E_Cell = 0) is connected to copper half-cell, the copper with be relatively less negative or less number of electrons.

$E_{(C u^{2 +} / C u)}^{o}$ = + 0.34 V; $E_{(F e^{2 +} / F e)}^{o}$ = $-$ 0.41 V

$E_{(C o^{2 +} / C o)}^{o}$ = $-$ 0.28 V; $E_{(Z n^{2 +} / Z n)}^{o}$ = $-$ 0.76 V

$∴$ Electrode potential of Cu $E_{(C u^{2 +} / C u)}^{o}$ show positive value.

41. ⇒ (JEE Main 2020 (Online) 6th September Evening Slot )

For the given cell :

Cu(s) | Cu²⁺(C₁M) || Cu²⁺(C₂M) | Cu(s)

change in Gibbs energy ( $Δ$ G) is negative, if :

(A) C₂ = $\sqrt{2}$ C₁

(B) C₂ = $\frac{C_{1}}{\sqrt{2}}$

(C) C₁ = 2C₂

(D) C₁ = C₂

Explanation

Correct answer is A

Given $Δ$ G < 0

$∴$ -nFE_cell < 0

$\Rightarrow$ E_cell > 0

We know, E_cell = $E_{c e l l}^{0}$ - $\frac{R T}{2 F} \ln (\frac{C_{1}}{C_{2}})$

= 0 - $\frac{R T}{2 F} \ln (\frac{C_{1}}{C_{2}})$

$∴$ - $\frac{R T}{2 F} \ln (\frac{C_{1}}{C_{2}})$ > 0

$\Rightarrow$ $\ln (\frac{C_{1}}{C_{2}})$ < 0

$\Rightarrow$ C₁ < C₂

By checking option, we can see

C₂ = $\sqrt{2}$ C₁ satisfy the condition C₁ < C₂.

42. ⇒ (JEE Main 2020 (Online) 4th September Morning Slot )

JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Electrochemistry Question 79 English
$E_{C u^{2 +} | C u}^{0}$ = +0.34 V

$E_{Z n^{2 +} | Z n}^{0}$ = -0.76 V

Identify the incorrect statement from the option below for the above cell :

(A) If E_ext < 1.1 V, Zn dissolves at anode and Cu deposits at cathode

(B) If E_ext = 1.1 V, no flow of e^– or current occurs

(C) If E_ext > 1.1 V, e^– flows from Cu to Zn

(D) If E_ext > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode

Explanation

Correct answer is D

$E_{c e l l}^{0}$ = $E_{C u^{2 +} | C u}^{0}$ - $E_{Z n^{2 +} | Z n}^{0}$

= 0.34 – (–0.76)

= 1.10 V

If E_ext< 1.1 V then Zn dissolves at anode and copper deposits at Cathode.

If E_ext > 1.1V then Zn deposited at zinc electrodes and Cu deposits at Cu electrode.

⇒Electrochemistry

Topic 3 : Cells and Electrode Potential, Nernst Equation 4