Compound A used as a strong oxidizing agent is amphoteric in nature. It is part of lead storage batteries. Compound A is :

1. PbO₂

2. PbO

3. PbSO₄

4. Pb₃O₄

 Emf of the following cell at 298 K in V is x × 10^–2 . The cell is Zn|Zn²⁺ (0.1 M) || Ag⁺ (0.01 M) | Ag. The value of x is-
(Rounded off to the nearest integer)
\(\begin{aligned} & \text { Given; } \mathrm{E}_{\mathrm{Zn}^{2+}}^{\mathrm{o}} / \mathrm{Zn}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=+0.80 \mathrm{~V} ; \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \end{aligned}\)

1. 157
2. 147
3. 144
4. 154

The gibbs energy change (in J) for the given reaction at  [Cu²⁺] = [Sn²⁺] = 1 M and 298K is-

$\mathrm{Cu}\left(\mathrm{s}\right) + {\mathrm{Sn}}^{2+} (\mathrm{aq}.) \to {\mathrm{Cu}}^{2+} (\mathrm{aq}.) + \mathrm{Sn}\left(\mathrm{s}\right)$
${\mathrm{E}}_{{\mathrm{sn}}^{2+}|\mathrm{Sn}}^0 = -0.16 \mathrm{V}, {\mathrm{E}}_{{\mathrm{Cu}}^{2+}|\mathrm{Cu}}^0 = 0.34 \mathrm{V} ,$
$\mathrm{Take} \mathrm{F}=96500 \mathrm{C} {\mathrm{mol}}^{-1}$

1. 97850 J

2. 3500 J

3. 45660 J

4. 96500 J

An oxidation-reduction reaction in which 3 electrons are transferred has a $∆G°$ of 17.37 kJ mol^–1 at 25°C. The value of $E_{cell}^o$(in V) is A× 10^–2. The value of A is-
(1 F = 96,500 C mol^–1)
1. -6
2. 4
3. -8
4. 2

The variation of molar conductivity with the concentration of an electrolyte (X) in an aqueous solution is shown in the given figure.

The electrolyte X is:

For the given cell :

$\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\left({\mathrm{C}}_1\mathrm{M}\right)\right|\left|{\mathrm{Cu}}^{2+}\left({\mathrm{C}}_2\mathrm{M}\right)\right|\mathrm{Cu}\left(\mathrm{s}\right)$ change in Gibbs energy $\left(∆\mathrm{G}\right)$ is negative, if:

1. ${\mathrm{C}}_1=2{\mathrm{C}}_2$

2. ${\mathrm{C}}_2=\raisebox{1ex}{${\mathrm{C}}_1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$

3. ${\mathrm{C}}_1={\mathrm{C}}_2$

4. ${\mathrm{C}}_2=\sqrt{2}{\mathrm{C}}_1$

A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using 0.1 Faraday electricity. The number of moles of Ni that will be deposited at the cathode are:

The standard Gibbs energy for the given cell reaction in kJ mol^–1 at 298 K is :
$Zn\left(s\right) + C{u}^{2+}\left(aq\right) \to Z{n}^{2+}\left(aq\right) + Cu\left(s\right)$
Eº(cell) = 2V at 298 K
(Faraday’s constant, F = 96000 C mol^–1)

Consider the values of reduction potential:
\(\mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+} ; E^{\circ}=+1.81 \mathrm{~V}\)
\(\mathrm{~Pb}^{4+}+2 e^{-} \rightarrow \mathrm{Pb}^{2+} ; E^{\circ}=+1.67 \mathrm{~V}\)
\(\mathrm{Ce}^{4+}+e^{-} \rightarrow C e^{3+} ; E^{\circ}=+1.61 \mathrm{~V}\)
\( \mathrm{Bi}^{3+}+3 e^{-} \rightarrow \mathrm{Bi} ; E^{\circ}=+0.20 \mathrm{~V}\)

The oxidizing power of the species will increase in the order of: