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To find the standard potential of M³⁺/M electrode, the following cell is constituted: Pt/M/M³⁺(0.001 mol L⁻¹⁾/Ag⁺(0.01 mol L−1)/Ag \ The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M³⁺+3e⁻ → M at 298 K will be : Given E_Ag⁺/Ag at 298 K=0.80 Volte
Option: 1 0.38 Volt
Option: 2 0.32 Volt
Option: 3 1.28 Volt
Option: 4 0.66 Volt

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Posted by

vishal kumar

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

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

$x = +1$

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

(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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Posted by

Kuldeep Maurya

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Given that the standard potentials (E^o) of Cu²⁺|Cu and Cu⁺|Cu are 0.34V and 0.522V respectively, the E^o of Cu²⁺|Cu⁺ is:
Option: 1 $-0.182 V$
Option: 2 $+0.158V$
Option: 3 $-0.158 V$
Option: 4 $0.182 V$

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.

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Posted by

Kuldeep Maurya

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 $2.1544$
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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Posted by

Kuldeep Maurya

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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 : $Cu(s)+Sn^{2+}(aq.)\rightarrow Cu^{2+}(aq.)+Sn(s);$ $\left ( E_{Sn^{2+}\mid Sn}^{0}=-0.16\; V, E_{Cu^{2+}\mid Cu}^{0}=0.34 V, \text {Take F=96500 C mol} ^{-1}\right )$

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

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

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

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

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Posted by

avinash.dongre

The equation that is incorrect is:
Option: 1 $(\Lambda ^{^{\circ}}_{m})_{NaBr}-(\Lambda ^{^{\circ}}_{m})_{NaI}= (\Lambda ^{^{\circ}}_{m})_{KBr}-(\Lambda ^{^{\circ}}_{m})_{NaBr}$
Option: 2 $(\Lambda ^{^{\circ}}_{m})_{NaBr}-(\Lambda ^{^{\circ}}_{m})_{NaCl}=(\Lambda ^{^{\circ}}_{m})_{KBr}-(\Lambda ^{^{\circ}}_{m})_{KCl}$
Option: 3 $(\Lambda ^{^{\circ}}_{m})_{NaBr}-(\Lambda ^{^{\circ}}_{m})_{NaCl}=(\Lambda ^{^{\circ}}_{m})_{KBr}-(\Lambda ^{^{\circ}}_{m})_{KCl}$
Option: 4 $(\Lambda ^{^{\circ}}_{m})_{H_{2}O}=(\Lambda ^{^{\circ}}_{m})_{HCl}+(\Lambda ^{^{\circ}}_{m})_{NaOH}-(\Lambda ^{^{\circ}}_{m})_{NaCl}$

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