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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\theta 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

What is the standard reduction potential \small \left ( E^{\circ} \right )  for Fe3+ \small \rightarrow Fe ? Given that: Fe2+ + 2e- \small \rightarrow Fe; \small E^{\circ} Fe2+ /Fe = -0.47V Fe3+ +e- \small \rightarrow Fe2+;\small E^{\circ} Fe3+/Fe2+=+0.77V  
Option: 1  −0.057 V
Option: 2 +0.057 V
Option: 3 +0.30 V
Option: 4  −0.30 V  
 

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Posted by

vishal kumar

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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

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 -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

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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

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

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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}} \\ \\

 

Both sides are not equal.

Therefore, Option(1) is correct.

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Posted by

Kuldeep Maurya

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.

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Posted by

Kuldeep Maurya

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The redox reaction among the following is:
Option: 1 Formation of ozone from atmospheric oxygen in presence of sunlight
Option: 2 Redox of H_{2}SO_{4} with NaOH 
Option: 3 Combination of dinitrogen with dioxygen of 2000K.
Option: 4 Reaction of [Co(H_{2}O)_{6} ]Cl_{3}  with AgNO_{3}
 

showing redox reaction, both oxidation and reduction.

Therefore, Option(3) is correct.

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Posted by

Ritika Jonwal

The oxidation states of iron atoms in compounds (A), (B) and (C), respectively,  are x, y and z. The sum of x, y and z is______.
 

 

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Posted by

Deependra Verma

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