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

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

If the Emf of the following cell Zn|Zn²⁺ (0.1 M) || Ag⁺ (0.01 M) | Ag at 298 K in V is \(x × 10^{–2}\). Find the value of x is:
(Rounded off to the nearest integer)

\(\begin{aligned} & \text {Given:} ~\mathrm{E}_{{\mathrm{Zn}^{2+}}/ \mathrm{Zn}}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=+0.80 \mathrm{~V} ; \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \end{aligned}\)

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

Given below are the half-cell reactions :

$\begin{array}{l}{\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn};{\mathrm{E}}^{\circ }=-1.18\mathrm{V}\\ 2({\mathrm{Mn}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+});{\mathrm{E}}^{\circ }=+1.51\mathrm{V}\end{array}$

The Eº for 3Mn²⁺ → Mn + 2Mn³⁺ will be :

1. – 2.69 V; the reaction will occur

2. – 0.33 V; the reaction will not occur

3. – 0.33 V; the reaction will occur

4. – 2.69 V; the reaction will not occur

The reduction potential of hydrogen half-cell will be negative if:

1. P(H₂) = 1atm and [H⁺] = 2.0 M

2. P(H₂) = 1 atm and [H⁺] = 1.0 M

3. P(H₂) = 2 atm and [H⁺] = 1.0 M

4. P(H₂) = 2 atm and [H⁺] = 2.0 M

Resistance of 0.2 M solution of an electrolyte is 50 $\mathrm{\Omega }$. The specific conductance of the solution is 1.3 S m^-1. If the resistance of the 0.4 M solution of the same electrolyte is \(260 \Omega,\) its molar conductivity is:
1. $6.25\times {10}^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$
2. $625\times {10}^{-4}{\mathrm{Sm}}^2{\mathrm{mol}}^{-1}$
3. $62.5\mathrm{S}{\mathrm{m}}^2{\mathrm{mol}}^{-1}$
4. $6250\mathrm{S}{\mathrm{m}}^2{\mathrm{mol}}^{-1}$

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

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

1. 97850 J

2. 3500 J

3. 45660 J

4. 96500 J

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: