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#### The oxidation number of potassium in $\mathrm{K_2O , K_2O_2 \: and\: KO_2}$ respectively is : Option: 1 Option: 2 Option: 3 Option: 4

$\mathrm{For \: K_2O:} 2x - 2 = 0\\\: \: \:$

$x = +1$

$\mathrm{For\ K_2O_2:} 2x - 2 = 0$

(In peroxide, the oxidation state of oxygen is -1)
$x = +1$

$\mathrm{For\ KO_2:} x - 2\times \frac{1}{2} = 0$

(In superoxide, the oxidation state of oxygen is -1/2)
$x = +1$

Therefore, Option(1) is correct.

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#### Given that the standard potentials (Eo) of Cu2+|Cu and Cu+|Cu are 0.34V and 0.522V respectively, the Eo of Cu2+|Cu+ is: Option: 1 Option: 2 Option: 3 Option: 4

Given,

$\mathrm{Cu^{2+}+2e^-\longrightarrow Cu, E^0 = 0.34~V} \quad (1)$

$\mathrm{Cu^{+}+e^-\longrightarrow Cu, E^0 = 0.522~V} \quad (2)$

We need to find out the electrode potential of the reaction

$\mathrm{Cu^{2+}+e^-\longrightarrow Cu^+, E^0 = ?} \quad (3)$

Reaction (3) can be obtained by carrying out the operation (1) - (2)

Thus, we can write

$\mathrm{\Delta G_3^0 =\Delta G_1^0 -\Delta G_2^0}$

$\Rightarrow\mathrm{-FE^0_3 = -2F(0.34)-(-F(0.522))}$

$\Rightarrow\mathrm{E^0_3 = 2(0.34)-(0.522)}$

$\Rightarrow\mathrm{E^0_3 = 0.158~V}$

Therefore, Option(2) is correct.

#### For an electrochemical cell $\mathrm{Sn(s)\left | Sn^{2+}(aq,1M) \right |\left | Pb^{2+}(aq,1M) \right |Pb(s)}$ the ratio $\mathrm{\frac{\left [ Sn^{2+} \right ]}{\left [ Pb^{2+} \right ]}}$ When this cell attains equilibrium is _____. $\mathrm{( Given :E^{0}_{Sn^{2+} | Sn} =-0.14V,}$ $\mathrm{E^{0}_{Pb^{2+} | Pb }=-0.13V,}$ $\mathrm{\frac{2.303RT}{F}=0.06)}$ Option: 1 Option: 2 1.11 Option: 3 7.15 Option: 4 3.14

As we have learnt,

Nernst equation is given as

$\mathrm{E= E^0_{cell}-\frac{2.303RT}{nF}logQ}$

Now, the chemical reaction occuring in the cell is given as

$\begin{array}{l}{\mathrm{Sn}+\mathrm{Pb}^{2+} \longrightarrow \mathrm{Sn}^{2+}+\mathrm{Pb}} \\ {0=0.01-\frac{0.06}{2} \log \left\{\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\}} \\ {0.01=\frac{0.06}{2} \log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\}} \\ {\frac{1}{3}=\log \left[\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}\right\} \Rightarrow \frac{\left[\mathrm{Sb}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}=10^{1 / 3}=2.1544}\end{array}$

Hence, the option number (1) is correct.

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#### The Gibbs energy change (in J) for the given reaction at $\left [ Cu^{2+} \right ]=\left [ Sn^{2+} \right ]=1 \; M$ and 298 K is : $Cu(s)+Sn^{2+}(aq.)\rightarrow Cu^{2+}(aq.)+Sn(s);$ $\left ( E_{Sn^{2+}\mid Sn}^{0}=-0.16\; V, E_{Cu^{2+}\mid Cu}^{0}=0.34 V, \text {Take F=96500 C mol} ^{-1}\right )$

$\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{+2}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{+2}(\mathrm{aq})+\mathrm{Sn}(\mathrm{s})$

\begin{aligned} \mathrm{E}^{0} &=-0.16-0.34 \\ &=-0.50 \end{aligned}

\begin{aligned} \Delta G^{0} &=-n F E^{0} \\ &=-2 \times 96500 \times(-0.5) \\ &=+96500 \end{aligned}

\begin{aligned} \Delta G &=\Delta G^{0}+R T \ell n Q \\ &=96500+\frac{25}{3} \times 298 \times 2.303 \log (1) \\ \Delta G &=96500 \text { Joules } \end{aligned}

#### The equation that is incorrect is: Option: 1 Option: 2 Option: 3 Option: 4

$\\\mathrm{(\Lambda ^0_m)_{NaBr}-(\Lambda ^0_m)_{NaI}= (\Lambda ^0_m)_{KBr}-(\Lambda ^0_m)_{NaBr}} \\\\\mathrm{(\Lambda ^0_m)_{Na}+(\Lambda ^0_m)_{Br}- (\Lambda ^0_m)_{Na}-(\Lambda ^0_m)_{I}= (\Lambda ^0_m)_{K}+(\Lambda ^0_m)_{Br} -(\Lambda ^0_m)_{Na}+(\Lambda ^0_m)_{Br}} \\ \\$

Both sides are not equal.

Therefore, Option(1) is correct.

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#### Potassium chlorate is prepared by the electrolysis of $KCl$ in basic solution $\mathrm{Cl}^{-}+6 \mathrm{OH}^{-} \rightarrow \mathrm{ClO}_{3}^{-}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}$ If only 60% of current utilized in the reaction, the time (hr) required to produce 10 g of $KClO_{3}$ using a current of 2A is ___ $\left(F=96500 \mathrm{C} \mathrm{mol}^{-},\: \mathrm{KClO}_{3}=122 \mathrm{g} \cdot \mathrm{mol}^{-}\right)$

The reaction occurs as follows:

$\mathrm{Cl^{-}\: +\: 6OH^{-}\: \rightarrow \: ClO_{3}^{-}\: +\: 3H_{2}O\: +\: 6e^{-}}$

Current efficiency = 60%

Now, moles of $KClO_{3}= \frac{10}{122.5}$

Thus, moles of $e^{-}= \frac{60}{122.5}$ (assuming 100% efficiency)

Thus, actual moles of $e^{-}= \frac{100}{122.5}$

$\mathrm{\therefore \: \frac{2\: x\: t}{96500}\: =\: \frac{100}{122.5}}$

Thus, t = 39387.76 sec. = 10.94 hours.

Thus, time required is approximately 11 hours.

#### The redox reaction among the following is: Option: 1 Formation of ozone from atmospheric oxygen in presence of sunlight Option: 2 Redox of  with   Option: 3 Combination of dinitrogen with dioxygen of 2000K. Option: 4 Reaction of   with

showing redox reaction, both oxidation and reduction.

Therefore, Option(3) is correct.

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#### For the given cell; $Cu(s)Cu^{2+}(C_{1}M)\parallel Cu^{2+}(C_{2}M)\mid Cu(s)$ change in Gibbs energy $(\Delta G)$ is negative, if :  Option: 1Option: 2Option: 3Option: 4

We know this formula

Only  following the above condition.

Therefore, the correct option is (4).