1. ⇒ (NEET 2023 )

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:

Assertion A : In equation $Δ_{rG} = - {nFE}_{cell}$ , value of $Δ_{rG}$ depends on n.

Reason R : $E_{cell}$ is an intensive property and $Δ_{rG}$ is an extensive property.

In the light of the above statements, choose the correct answer from the options given below:

(A) Both A and R are true but R is NOT the correct explanation of A.

(B) A is true but R is false.

(C) A is false but R is true.

(D) Both A and R are true and R is the correct explanation of A.

Correct Answer is Option (D)

$Δ_{rG} = - {nFE}_{cell}$

$E_{cell}$ is an intensive property and $Δ_{r} G$ is an extensive property as it depends on number of $e^{Θ}$ transferred in cell reaction.

2. ⇒ (NEET 2022 Phase 1 )

At 298 K, the standard electrode potentials of Cu²⁺ / Cu, Zn²⁺ / Zn, Fe²⁺ / Fe and Ag⁺ / Ag are 0.34 V, $-$ 0.76 V, $-$ 0.44 V V and 0.80 V, respectively.

On the basis of standard electrode potential, predict which of the following reaction cannot occur?

(A) CuSO₄(aq) + Zn(s) $\to$ ZnSO₄(aq) + Cu(s)

(B) CuSO₄(aq) + Fe(s) $\to$ FeSO₄(aq) + Cu(s)

(C) FeSO₄(aq) + Zn(s) $\to$ ZnSO₄(aq) + Fe(s)

(D) 2CuSO₄(aq) + 2Ag(s) $\to$ 2Cu(s) + Ag₂SO₄(aq)

Correct Answer is Option (D)

For a reaction to be spontaneous, E $_{c e l l}^{o}$ must be positive.

$∙$ For, FeSO₄(aq) + Zn(s) $\to$ ZnSO₄(aq) + Fe(s)

E $_{c e l l}^{o}$ = E $_{c a t h o d e}^{o}$ $-$ E $_{a n o d e}^{o}$

= $-$ 0.44 V $-$ ( $-$ 0.76 V)

= 0.32 V

$∙$ For, 2CuSO₄(aq) + 2Ag(s) $\to$ 2Cu(s) + Ag₂SO₄(aq)

E $_{c e l l}^{o}$ = 0.34 V $-$ 0.80 V

= $-$ 0.46 V

$∙$ For, CuSO₄(aq) + Zn(s) $\to$ ZnSO₄(aq) + Cu(s)

E $_{c e l l}^{o}$ = 0.34 V $-$ ( $-$ 0.76 V)

= 1.1 V

$∙$ For, CuSO₄(aq) + Fe(s) $\to$ FeSO₄(aq) + Cu(s)

E $_{c e l l}^{o}$ = 0.80 V $-$ ( $-$ 0.44 V)

= 1.24 V

3. ⇒ (NEET 2022 Phase 1 )

Given below are half cell reactions:

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ ,

$E_{M n^{2 +} / M n O_{4}^{-}}^{o} = - 1.510 V$

$\frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-} \to H_{2} O$

$E_{O_{2} / H_{2} O}^{o} = + 1.223 V$

Will the permanganate ion, $M n O_{4}^{-}$ liberate O₂ from water in the presence of an acid?

(A) Yes, because $E_{c e l l}^{o}$ = + 0.287 V

(B) No, because $E_{c e l l}^{o}$ = $-$ 0.287 V

(C) Yes, because $E_{c e l l}^{o}$ = + 2.733 V

(D) No, because $E_{c e l l}^{o}$ = $-$ 2.733 V

Correct Answer is Option (A)

$∙$ $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ ..... (i)

$E_{M n O_{4}^{-} / M n^{2 +}}^{o} = - E_{M n^{2 +} / M n O_{4}^{-}}^{o} = 1.51 V$

$∙$ $H_{2} O \to \frac{1}{2} O_{2} + 2 H^{+} + 2 e^{-}$ ..... (ii)

$E_{O_{2} / H_{2} O}^{o} = 1.223 V$

Using 2 $\times$ (i) + 5 $\times$ (ii), net cell reactions is

$2 M n O_{4}^{-} + 6 H^{+} \to 2 M n^{2 +} + \frac{5}{2} O_{2} + 3 H_{2} O$

$E_{c e l l}^{o} = E_{C}^{o} - E_{A}^{o} = E_{M n O_{4}^{-} / M n^{2 +}}^{o} - E_{O_{2} / H_{2} O}^{o} = 1.51 - 1.223 = 0.287 V$

Since $E_{c e l l}^{o} > 0$ , therefore net cell reaction is spontaneous and so $M n O_{4}^{-}$ liberate O₂ from H₂O in presence of an acid.

4. ⇒ (NEET 2022 Phase 1 )

Find the emf of the cell in which the following reaction takes place at 298 K

Ni(s) + 2Ag⁺ (0.001 M) $\to$ Ni²⁺ (0.001 M) + 2Ag(s)

(Given that E $_{c e l l}^{o}$ = 10.5 V, $\frac{2.303 R T}{F} = 0.059$ at 298 K)

(A) 1.0385 V

(B) 1.385 V

(C) 0.9615 V

(D) 1.05 V

Correct Answer is Option (A)

Ni(s) + 2Ag⁺ (0.001 M) $\to$ Ni²⁺ (0.001 M) + 2Ag(s)

E $_{c e l l}^{o}$ = 10.5 V

E_cell = E $_{c e l l}^{o}$ $-$ $\frac{0.059}{n} \log \frac{[N i^{2 +}]}{{[A g^{+}]}^{2}}$

$= 10.5 - \frac{0.059}{2} \log \frac{(10^{- 3})}{{(10^{- 3})}^{2}}$

$\Rightarrow 10.5 - \frac{0.059}{2} \log (10)^{3}$

$\Rightarrow 10.5 - 0.0295 \times 3$

$= 10.5 - 0.0885$

$= 10.4115$ V

5. ⇒ (NEET 2019 )

For a cell involving one electron $E_{c e l l}^{Θ}$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is :

[Given that $\frac{2.303 R T}{F}$ = 0.059 V at T = 298 K ]

(A) 1.0 $\times$ 10³⁰

(B) 1.0 $\times$ 10¹⁰

(C) 1.0 $\times$ 10²

(D) 1.0 $\times$ 10⁵

Correct Answer is Option (B)

We know,

E_cell = $E_{c e l l}^{Θ}$ - $\frac{2.303 R T}{n F} \log Q$

$\Rightarrow$ E_cell = $E_{c e l l}^{Θ}$ - $\frac{0.059}{n} \log Q$

(At equilibrium, E_cell = 0 and Q = K_eq)

$\Rightarrow$ 0 = $E_{c e l l}^{Θ}$ - $\frac{0.059}{1} \log K_{e q}$

$\Rightarrow$ $\log K_{e q} = \frac{E_{c e l l}^{Θ}}{0.059}$ = $\frac{0.59}{0.059}$ = 10

$\Rightarrow$ K_eq = 10¹⁰

6. ⇒ (NEET 2019 )

For the cell reaction
2Fe³⁺(aq) + 2I^– (aq) $\to$ 2Fe²⁺(aq) + I₂(aq)
$E_{c e l l}^{Θ}$ = 0.24 V at 298 K. The standard Gibbs energy ( $Δ$ _rG^o) of the cell reaction is :
[Given that Faraday constant F = 96500 C mol^–1]

(A) 46.32 kJ mol^–1

(B) 23.16 kJ mol^–1

(C) –46.32 kJ mol^–1

(D) –23.16 kJ mol^–1

Correct Answer is Option (C)

Here, n = 2

$Δ$ G^o = -nFE^o

= – 2 × 96500 × 0.24

= – 46320 J mol^–1

= – 46.32 KJ mol^–1

7. ⇒ (NEET 2017 )

In the electrochemical cell :
$Z n | Z n S O_{4} (0.01 M) |$ $| C u S O_{4} (1.0 M) | C u,$
the emf of this Daniell cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E₂. From the followings, which one is the relationship between E₁ and E₂? (Given, RT/F = 0.059)

(A) E₁ < E₂

(B) E₁ > E₂

(C) E₂ = 0¹E₁

(D) E₁ = E₂

Correct Answer is Option (B)

E_cell = E^o - $\frac{0.059}{n} \log \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

E₁ = E^o - $\frac{0.059}{2} \log \frac{0.01}{1}$

= E^o - $\frac{0.059}{2} (- 2) \log 10$

E₁ = E^o + 0.059

E₂ = E^o - $\frac{0.059}{2} \log \frac{1}{0.01}$

$\Rightarrow$ E₂ = E^o - $\frac{0.059}{2} \log 100$

$\Rightarrow$ E₂ = E^o - 0.059

$∴$ E₁ > E₂

8. ⇒ (NEET 2016 Phase 2 )

If the E^o_cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of $Δ$ G^o and K_eq ?

(A) $Δ$ G^o > 0; K_eq < 1

(B) $Δ$ G^o > 0; K_eq > 1

(C) $Δ$ G^o < 0; K_eq > 1

(D) $Δ$ G^o < 0; K_eq < 1

Correct Answer is Option (A)

We know that

$Δ$ G^o = –nFE^o_cell

E^o_cell = -ve then $Δ$ G^o = +ve or $Δ$ G^o > 0

$Δ$ G^o = -nRTlog K_eq

For $Δ$ G^o = +ve, K_eq = -ve or K_eq < 1.

9. ⇒ (2015)

The pressure of H₂ required to make the potential of H₂-electrode zero in pure water at 298 K is

(A) 10^{$-$ 10} atm

(B) 10^{$-$ 4} atm

(C) 10^{$-$ 14} atm

(D) 10^{$-$ 12} atm

Correct Answer is Option (C)

2H⁺ (aq) + 2e^– $\to$ H₂(g)

E_cell = E^o_cell - $\frac{0.0591}{2} \log \frac{p_{H_{2}}}{{[H^{+}]}^{2}}$

$\Rightarrow$ 0 = 0 - $\frac{0.0591}{2} \log \frac{p_{H_{2}}}{{[10^{- 7}]}^{2}}$

$\Rightarrow$ $\frac{p_{H_{2}}}{{[10^{- 7}]}^{2}} = 1$

$\Rightarrow$ P_H₂ = 10^–14 atm

10. ⇒ (NEET 2013 (Karnataka) )

Consider the half-cell reduction reaction
Mn²⁺ + 2e^$-$ $\to$ Mn, E^o = $-$ 1.18 V
Mn²⁺ $\to$ Mn³⁺ + e^$-$, E^o = $-$ 1.51 V
The $E$ ^o for the reaction 3 Mn²⁺ $\to$ Mn^o + 2Mn³⁺, and possibility of the forward reaction are respectively

(A) $-$ 4.18 V and yes

(B) + 0.33 V and yes

(C) + 2.69 V and no

(D) $-$ 2.69 V and no

Correct Answer is Option (D)

Mn²⁺ + 2e^$-$ $\to$ Mn, E^o = $-$ 1.18 V

Mn²⁺ $\to$ Mn³⁺ + e^$-$, E^o = $-$ 1.51 V

3Mn²⁺ $\to$ Mn^o + 2Mn³⁺, $Δ$ E^o = - 1.81 - 1.51 = $-$ 2.69 V

Since $Δ$ E° is negative,

$∴$ $Δ$ G = –nFE°, $Δ$ G will have positive value so, forward reaction is not possible.

⇒Electrochemistry

Topic 2 : Nernst Equation 1