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JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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The equation representing kohlrausch law from the following is (V+={V^ + } = No. of cations, V−={V^ - } = No. of anions)
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Correct option: B
λmo=V+λ+o+V−λ−o\lambda _m^\mathrm{o} = {V^ + }\lambda _ + ^\mathrm{o} + {V^ - }\lambda _ - ^\mathrm{o}
Option B (λmo=V+λ+o+V−λ−o\lambda _m^\mathrm{o} = {V^ + }\lambda _ + ^\mathrm{o} + {V^ - }\lambda _ - ^\mathrm{o}) is correct because it represents Kohlrausch's Law, stating that the limiting molar conductivity of an electrolyte is the sum of the contributions of the individual ions.

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At 25∘C25^{\circ} \mathrm{C}, the standard emf of cell having reactions involving two electron change is found to be 0.295 V0.295 \mathrm{~V}. The equilibrium constant of the reaction is Which are used as secondary reference electrodes? For the following cell with hydrogen electrodes at two different pressures p1andp2{p_1}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {p_2}

Pt(H2)p1∣H+(aq)1M∣Pt(H2)p2\mathop {\mathrm{Pt}\left( {{\mathrm{H}_2}} \right)}\limits_{{p_1}} \left| {\mathop {{\mathrm{H}^ + }\left( {aq} \right)}\limits_{1\mathrm{ }M} } \right|\mathop {Pt\left( {{\mathrm{H}_2}} \right)}\limits_{{p_2}}

emf is given by
A current of strength 2.5 A was passed through CuSO4\mathrm{CuS}{\mathrm{O}_4} solution for 6 min 26 s. The amount of copper deposited is (At. Wt. of Cu = 63.5, 1F = 96500 C) A current strength of 9.65 amperes is passed through excess fused AlCl3{AlCl}_3 for 5 hours. How many litres of chlorine will be liberated at STPSTP? (F=96500C)({F}=96500 {C}) The electrode potential (in millivolts) of 2Ag(s)+S2−(aq)→Ag2S(s)+2e−2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Ag}_2 \mathrm{S}(\mathrm{s})+2 \mathrm{e}^{-} in a solution buffered at pH=3\mathrm{pH}=3 and which is also saturated with 0.1 M−H2S,0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S}, is (Given: for H2S,Ka1=10−8;Ka2=10−13,Ksp\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}} of Ag2S=4×10−48,\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48}, EAg∣Ag+∘=0.80 V,E^{\circ}_{{\mathrm{Ag} | \mathrm{Ag}}^{+}}=0.80 \mathrm{~V}, log⁡2=0.3,\log 2=0.3, 2.303RT/F=0.062.303 R T / F=0.06)
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