1. ⇒ (NEET 2023 Manipur )

The $E^{Θ}$ values for

$\begin{aligned} {Al}^{+} / Al = + 0.55 V and {Tl}^{+} / Tl = - 0.34 V \\ {Al}^{3 +} / Al = - 1.66 V and T^{3 +} / Tl = + 1.26 V \end{aligned}$

Identify the incorrect statement

(A) $Al$ is more electropositive than $Tl$

(B) ${Tl}^{3 +}$ is a good reducing agent than ${Tl}^{1 +}$

(C) ${Al}^{+}$ is unstable in solution

(D) $Tl$ can be easily oxidised to ${Tl}^{+}$ than ${Tl}^{3 +}$

Correct Answer is Option (B)

${Tl}^{3}$ act as an oxidising agent not reducing agent.

2. ⇒ (NEET 2023 Manipur )

The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is

$\begin{aligned} {Fe}_{(aq)}^{2 +} + 2 e^{-} \to Fe (s), E^{\circ} = - 0.44 V \\ {Cr}_{2} O_{7}^{2 -} (aq) + 14 H^{+} + 6 e^{-} \to 2 {Cr}^{3 +} + 7 H_{2} O \\ E^{\circ} = + 1.33 V \end{aligned}$

(A) $+ 1.77 V$

(B) $+ 2.65 V$

(C) $+ 0.01 V$

(D) $+ 0.89 V$

Correct Answer is Option (A)

$\begin{aligned} E_{cell}^{\circ} & = E_{C}^{o} - E_{A}^{o} \\ = (1.33) - (- 0.44) \\ = + 1.77 V \end{aligned}$

3. ⇒ (NEET 2022 Phase 2 )

Two half cell reactions are given below.

$C o^{3 +} + e^{-} \to C o^{2 +}, E_{C o^{2 +} / C o^{3 +}}^{0} = - 1.81 V$

$2 A l^{3 +} + 6 e^{-} \to 2 A l (s), E_{A l / A l^{3 +}}^{0} = + 1.66 V$

The standard EMF of a cell with feasible redox reaction will be :

(A) $-$ 3.47 V

(B) +7.09 V

(C) +0.15 V

(D) +3.47 V

Correct Answer is Option (D)

Since $E_{O P}^{0}$ of Al is more than Co²⁺, so at anode Al will oxidise and at cathode Co³⁺ will reduce.

$E_{C e l l}^{0} = (E_{C a t h o d e}^{0})_{R P} - (E_{A n o d e}^{0})_{R P}$

$= E_{C o^{3 +} / C o^{2 +}}^{0} - E_{A l^{3 +} / A l}^{0}$

$= (1.81) - (- 1.66)$

$= + 3.47 V$

4. ⇒ (NEET 2016 Phase 2 )

Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because

(A) zinc is lighter than iron

(B) zinc has lower melting point than iron

(C) zinc has lower negative electrode potential than iron

(D) zinc has higher negative electrode potential than iron

Correct Answer is Option (D)

The reduction potential values are

E°_Zn²⁺/Zn = – 0.76 V

E°_Fe²⁺/Fe = – 0.44 V

Thus, due to higher negative electro potential value of zinc than iron, iron cannot be coated on zinc.

5. ⇒ (NEET 2013 )

A button cell used in watches function as following.
Zn_(s) + Ag₂O_(s) + H₂O_(l) $⇌$ 2Ag_(s) + Zn²⁺_(aq) + 2OH^$-$_(aq)

If half cell potentials are
Zn²⁺_(aq) + 2e^$-$ $\to$ Zn_(s); E^o = $-$ 0.76 V
Ag₂O_(s) + H₂O_(l) + 2e^$-$ $\to$ 2Ag_(s) + 2OH^$-$_(aq), E^o = 0.34 V

The cell potential will be

(A) 0.84 V

(B) 1.34 V

(C) 1.10 V

(D) 0.42 V

Correct Answer is Option (C)

At anode :Zn²⁺_(aq) + 2e^$-$ $\to$ Zn_(s); E^o = $-$ 0.76 V

At cathode : Ag₂O_(s) + H₂O_(l) + 2e^$-$ $\to$ 2Ag_(s) + 2OH^$-$_(aq), E^o = 0.34 V

E°_cell = E°_cathode – E°_anode

= 0.34 – (–0.76) = 1.10 V

6. ⇒ (AIPMT 2012 Mains )

Standard reduction potentials of the half reactions are given below :
F_2(g) + 2e^$-$ $\to$ 2F^$-$_(aq) ;   E^o = + 2.85 V
Cl_2(g) + 2e^$-$ $\to$ 2Cl^$-$_(aq) ;   E^o = + 1.36 V
Br_2(l) + 2e^$-$ $\to$ 2Br^$-$_(aq) ;   E^o = + 1.06 V
I_2(s) + 2e^$-$ $\to$ 2I^$-$_(aq) ;  E^o = + 0.53 V

The strongest oxidising and reducing agents 23 respectively are

(A) F₂ and I^$-$

(B) Br₂ and Cl^$-$

(C) Cl₂ and Br^$-$

(D) Cl₂ and I₂

Correct Answer is Option (A)

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.

7. ⇒ (AIPMT 2011 Mains )

A solution contains Fe²⁺, Fe³⁺ and I^$-$ ions. This solution was treated with iodine at 35^oC. E^o for Fe³⁺/Fe²⁺ is + 0.77 V and E^o for I₂/2I^$-$ = 0.536 V.
The favourable redox reaction is

(A) I₂ will be reduced to I^$-$

(B) there will be no redox reaction

(C) I^$-$ will be oxidised to I₂

(D) Fe²⁺ will be oxidised to Fe³⁺

Correct Answer is Option (C)

Since the reduction potential of Fe³⁺/Fe²⁺ is greater than that of I₂ /I^– , Fe³⁺ will be reduced and I^– will be oxidised.

2Fe³⁺ + 2I^– $\to$ 2Fe²⁺ + I₂

8. ⇒ (AIPMT 2011 Prelims )

Standard electrode potential of three metals X, Y and Z are $-$ 1.2 V, + 0.5 V and $-$ 3.0 V respectively. The reducing power of these metals will be

(A) Y > Z > X

(B) Y > X > Z

(C) Z > X > Y

(D) X > Y > Z

Correct Answer is Option (C)

As the electrode potential drops, reducing power increases.

So, Z (–3.0 V) > X (–1.2 V) > Y (+ 0.5 V)

⇒Electrochemistry

Topic 1 : Galvanic Cell 1