11. ⇒ (JEE Main 2023 (Online) 31st January Morning Shift )

The logarithm of equilibrium constant for the reaction ${Pd}^{2 +} + 4 {Cl}^{-} ⇌ {PdCl}_{4}^{2 -}$ is ___________ (Nearest integer)

Given : $\frac{2.303 R T}{F} = 0.06 V$

${Pd}_{(aq)}^{2 +} + 2 e^{-} ⇌ Pd (s) E^{⊖} = 0.83 V$

$\begin{aligned} {PdCl}_{4}^{2 -} (aq) + 2 e^{-} ⇌ Pd (s) + 4 {Cl}^{-} (aq) E^{⊖} = 0.65 V \end{aligned}$

correct answer is 6

# Explanation

Sol. $Δ G^{\circ} = - R T ℓ n K$

$\begin{aligned} - {nFE}_{cell}^{o} = - RT \times 2.303 (\log_{10} K) \\ \frac{E_{Cell}^{o}}{0.06} \times n = \log K . . . . . . . (1) \\ {Pd}^{+ 2} (aq.) + {2 e}^{-} ⇌ Pd (s), E_{cat, r e d u c t i o n}^{o} = 0.83 \\ Pd (s) + 4 {Cl}^{-} (aq.) ⇌ {PdCl}_{4}^{2 -}, (aq) + 2 e^{-}, E_{Anode, Oxidation}^{0} = 0.65 \end{aligned}$

Net Reaction $\to {Pd}^{2 +}$ (aq.) $+ 4 {Cl}^{-}$ (aq.) $⇌ {PdCl}_{4}^{2 -}$ (aq.)

$\begin{aligned} E_{cell}^{0} = E_{cat,red}^{o} - E_{Anded, oxidid}^{0} \\ E_{cell}^{o} = 0.83 - 0.65 \\ E_{cell}^{0} = 0.18 . . . . . . . . . (2) \end{aligned}$

Also $n = 2$ .......(3)

Using equation (1), (2) and (3)

$\log K = 6$

12. ⇒ (JEE Main 2023 (Online) 30th January Evening Shift )

The electrode potential of the following half cell at

298 K

X | X^{2 +} (0.001 M) ‖ Y^{2 +} (0.01 M) | Y

is _______

\times 10^{- 2} V

(Nearest integer)

Given:

E_{X^{2 +} ∣ X}^{0} = - 2.36 V

E_{Y^{2 +} ∣ Y}^{0} = + 0.36 V

\frac{2.303 RT}{F} = 0.06 V

correct answer is 275

# Explanation

$\begin{aligned} Cell reaction : X + Y^{2 +} (0.01 M) ⟶ X^{2 +} (0.001 M) + Y \\ E_{cell}^{o} = 0.36 - (- 2.36) = 2.72 V \\ E_{cell} = E_{cell}^{\circ} - \frac{2.303 R T}{n F} \log \frac{[X^{2 +}]}{[Y^{2 +}]} \\ = 2.72 - \frac{0.06}{2} \log \frac{(0.001)}{0.01} [n = 2, as two electrons are involved in the reaction] \\ = 2.72 - (0.03) \times (- 1) \\ = 2.72 + 0.03 \\ = 2.75 V = 275 \times 10^{- 2} V \end{aligned}$

13. ⇒ (JEE Main 2023 (Online) 30th January Morning Shift )

Consider the cell

${Pt}_{(s)} | H_{2} (g, 1 atm) | H^{+} (aq, 1 M) | | {Fe}^{3 +} (aq), {Fe}^{2 +} (aq) ∣ Pt (s)$

When the potential of the cell is $0.712 V$ at $298 K$ , the ratio $[{Fe}^{2 +}] / [{Fe}^{3 +}]$ is _____________. (Nearest integer)

Given : ${Fe}^{3 +} + e^{-} = {Fe}^{2 +}, E^{θ} {Fe}^{3 +}, {Fe}^{2 +} ∣ Pt = 0.771$

$\frac{2.303 RT}{F} = 0.06 V$

correct answer is 10

# Explanation

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

Cathode $(F e^{3 +} + e^{-} \to F e^{2 +}) \times 2$

$H_{2} + 2 F e^{3 +} \to 2 H^{+} + 2 F e^{2 +}$

$E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{2} \log {(\frac{F e^{2 +}}{F e^{3 +}})}^{2}$

$0.712 = 0.771 - 0.059 \log \frac{F e^{2 +}}{F e^{3 +}}$

$- 0.059 = - 0.059 \log \frac{F e^{2 +}}{F e^{3 +}}$

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

14. ⇒ (JEE Main 2023 (Online) 29th January Evening Shift )

The equilibrium constant for the reaction

$Zn (s) + {Sn}^{2 +} (aq)$ $⇌$ ${Zn}^{2 +} (aq) + Sn (s)$ is $1 \times 10^{20}$ at 298 K. The magnitude of standard electrode potential of $Sn / {Sn}^{2 +}$ if $E_{Z n^{2 +} / Zn}^{Θ} = - 0.76 V$ is __________ $\times 10^{- 2}$ V. (Nearest integer)

Given : $\frac{2.303 RT}{F} = 0.059 V$

correct answer is 17

# Explanation

$E_{c e l l}^{o} = \frac{2.303 R T}{2 F} \log k$

$E_{c e l l}^{o} = \frac{0.059}{2} \log (10^{20})$

$E_{Z n^{2 +} / Z n}^{o} + 0.76 = 0.59$

$E_{Z n^{2 +} / Z n}^{o} = 0.59 - 0.76$

$E_{Z n / Z n^{2 +}}^{o} = 0.17 V$

15. ⇒ (JEE Main 2023 (Online) 25th January Evening Shift )

$P t (s) | H_{2} (g) (1 b a r) | H^{+} (a q) (1 M) | | M^{3 +} (a q), M^{+} (a q) | P t (s)$

The $E_{cell}$ for the given cell is 0.1115 V at 298 K when $\frac{[M^{+} (a q)]}{[M^{3 +} (a q)]} = 10^{a}$

The value of $a$ is ____________

Given : $E_{M^{3 +} / M^{+}}^{θ} = 0.2$ V

$\frac{2.303 R T}{F} = 0.059 V$

correct answer is 3

# Explanation

Overall reaction :-

$\begin{aligned} H_{2 (g)} + M_{(aq)}^{3 +} ⟶ M_{(aq)}^{+} + 2 H_{(aq)}^{+} \\ E_{Cell} = E_{Cathode}^{o} - E_{anode}^{o} - \frac{0.059}{2} \log \frac{[M^{+}] \times 1^{2}}{[M^{+ 3}] 1} \\ 0.1115 = 0.2 - \frac{0.059}{2} \log \frac{[M^{+}]}{[M^{+ 3}]} \\ 3 = \log \frac{[M^{+}]}{[M^{+ 3}]} \\ ∴ a = 3 \end{aligned}$

16. ⇒ (JEE Main 2023 (Online) 25th January Morning Shift )

Consider the cell

$Pt (s) | H_{2} (g) (1 atm) | H^{+} (aq, [H^{+}] = 1) | | F e^{3 +} (aq), F e^{2 +} (aq) | Pt (s)$

Given $E_{F e^{3 +} / F e^{2 +}}^{o} = 0.771 V$ and $E_{H^{+} / 1 / 2 H_{2}}^{o} = 0 V, T = 298 K$

If the potential of the cell is 0.712 V, the ratio of concentration of Fe $^{2 +}$ to Fe $^{3 +}$ is _____________ (Nearest integer)

correct answer is 10

# Explanation

$\begin{aligned} \frac{1}{2} H_{2} (g) + {Fe}^{3 +} (aq .) ⟶ H^{+} (aq) + {Fe}^{2 +} (aq .) \\ E = E^{o} - \frac{0.059}{1} \log \frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]} \\ \Rightarrow 0.712 = (0.771 - 0) - \frac{0.059}{1} \log \frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]} \\ \Rightarrow \log \frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]} = \frac{(0.771 - 0712)}{0.059} = 1 \\ \Rightarrow \frac{[{Fe}^{2 +}]}{[{Fe}^{3 +}]} = 10 \end{aligned}$

17. ⇒ (JEE Main 2023 (Online) 24th January Morning Shift )

At 298 K, a 1 litre solution containing 10 mmol of $C r_{2} O_{7}^{2 -}$ and 100 mmol of ${Cr}^{3 +}$ shows a pH of 3.0.

Given : $C r_{2} O_{7}^{2 -} \to C r^{3 +}; E^{\circ} = 1.330$ V

and $\frac{2.303 RT}{F} = 0.059$ V

The potential for the half cell reaction is $x \times 10^{- 3}$ V. The value of $x$ is __________

correct answer is 917

# Explanation

$\begin{aligned} {Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O \\ E = 1.33 - \frac{0.059}{6} \log \frac{(0.1)^{2}}{(10^{- 2}) {(10^{- 3})}^{14}} \\ E = 1.33 - \frac{0.059}{6} \times 42 = 0.917 \\ E = 917 \times 10^{- 3} \\ x = 917 \end{aligned}$

18. ⇒ (JEE Main 2022 (Online) 29th July Evening Shift )

For a cell, $Cu (s) | {Cu}^{2 +} (0.001 M) ‖ {Ag}^{+} (0.01 M) | Ag (s)$

the cell potential is found to be $0.43 V$ at $298 K$ . The magnitude of standard electrode potential for ${Cu}^{2 +} / Cu$ is _________ $\times 10^{- 2} V$ .

[Given : $E_{A g^{+} / A g}^{Θ}$ = 0.80 V and $\frac{2.303 R T}{F}$ = 0.06 V]

correct answer is 34

# Explanation

Anode : $Cu (s) ⟶ {Cu}^{2 +} (aq) + 2 e^{-}$

Cathode : $[{Ag}^{+} + e^{-} ⟶ Ag (s)] 2$

Cus(s) $+ 2 {Ag}^{+} (aq) ⟶ {Cu}^{2 +} (aq) + 2 Ag (s)$

$\begin{aligned} E_{cell} = E_{cell}^{0} - \frac{0.06}{2} \log \frac{[{Cu}^{2 +}]}{{[{Ag}^{+}]}^{2}} \\ 0.43 = E_{cell}^{0} - \frac{0.06}{2} \log (\frac{10^{- 3}}{{(10^{- 2})}^{2}}) \\ 0.43 = E_{cell}^{0} - 0.03 \log 10 \\ E_{cell}^{0} = 0.46 V \\ E_{cell}^{0} = E_{{Ag}^{+} / Ag}^{0} - E_{{Cu}^{2 +} / Cu}^{0} \\ E_{{Cu}^{2 +} / Cu}^{0} = (0.80 - 0.46) = 0.34 V = 34 \times 10^{- 2} \end{aligned}$

19. ⇒ (JEE Main 2022 (Online) 25th July Evening Shift )

The spin-only magnetic moment value of M³⁺ ion (in gaseous state) from the pairs Cr³⁺ / Cr²⁺, Mn³⁺ / Mn²⁺, Fe³⁺ / Fe²⁺ and Co³⁺ / Co²⁺ that has negative standard electrode potential, is ____________ B.M. [Nearest integer]

correct answer is 4

# Explanation

Among the pairs given, ${Cr}^{3 +} / {Cr}^{2 +}$ has negative reduction potential which is $- 0.41 V$ .

$Cr$ (III) $\Rightarrow d^{3}$

Number of unpaired electrons $= 3$

$μ = \sqrt{3 (3 + 2)} = \sqrt{15} ≃ 4$ B.M.

20. ⇒ (JEE Main 2022 (Online) 25th July Morning Shift )

The cell potential for $Zn | {Zn}^{2 +} (aq) | | {Sn}^{x +} | Sn$ is $0.801 V$ at $298 K$ . The reaction quotient for the above reaction is $10^{- 2}$ . The number of electrons involved in the given electrochemical cell reaction is ____________.

$($ Given $: E_{{Zn}^{2 +} ∣ Zn}^{o} = - 0.763 V, E_{{Sn}^{x +} ∣ Sn}^{o} = + 0.008 V$ and $\frac{2.303 RT}{F} = 0.06 V)$

correct answer is 4

# Explanation

$A : Zn \to {Zn}^{2 +} + 2 e^{-}$

$C : {Sn}^{+ x} + {xe}^{-} \to Sn$

$E_{Cell}^{\circ} = E_{Zn ∣ {Zn}^{2 +}}^{\circ} + E_{{Sn}^{+ x} ∣ Sn}^{\circ}$

$\Rightarrow 0.763 + 0.008 = 0.771 V$

From the Nernst equation,

$E_{Cell} = E_{Cell}^{\circ} \frac{- 2.303 RT}{nF} \log Q$

$0.801 = 0.771 - \frac{0.06}{n} \log 10^{- 2}$

$0.03 = \frac{0.06}{n} \times 2$

$n = 4$

21. ⇒ (JEE Main 2022 (Online) 30th June Morning Shift )

In which of the following half cells, electrochemical reaction is pH dependent?

(A) $P t | F e^{3 +}, F e^{2 +}$

(B) $M n O_{4}^{-} | M n^{2 +}$

(C) $A g | A g C l | C l^{- 1}$

(D) $\frac{1}{2} F_{2} | F^{-}$

Explanation

Correct answer is B

Reduction of ${MnO}_{4}^{-}$ is $pH$ dependent.

In acidic medium

${MnO}_{4}^{-} + 5 e^{-} ⟶ {Mn}^{2 +}$

In neutral medium

${MnO}_{4}^{-} + 3 e^{-} ⟶ {Mn}^{4 +}$

In basic medium

${MnO}_{4}^{-} + e ⟶ {Mn}^{6 +}$

So, according to $pH$ , the reaction and potential of cell changes.

⇒Electrochemistry

Topic 3 : Cells and Electrode Potential, Nernst Equation 2