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JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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How long a current of 3 ampere has to be passed through a solution of silver nitrate to coat a metal surface of 80 cm2\mathrm{cm}^2 with a 0.005Β mm0.005 \mathrm{~mm} thick layer ?Density of silver is 10.5Β g/cm310.5 \mathrm{~g} / \mathrm{cm}^3.
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Correct option: A
125
Option A (125) is correct because the time required is calculated using Faraday's laws of electrolysis, considering the volume, density, and current.

More Questions from ELECTRO CHEMISTRY

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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)+S2βˆ’(aq)β†’Ag2S(s)+2eβˆ’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 0.1Β Mβˆ’H2S,0.1 \mathrm{~M}-\mathrm{H}_2 \mathrm{S}, is (Given: for H2S,Ka1=10βˆ’8;Ka2=10βˆ’13,Ksp\mathrm{H}_2 \mathrm{S}, K_{\mathrm{a}_1}=10^{-8} ; K_{\mathrm{a}_2}=10^{-13}, K_{\mathrm{sp}} of Ag2S=4Γ—10βˆ’48,\mathrm{Ag}_2 \mathrm{S}=4 \times 10^{-48}, EAg∣Ag+∘=0.80Β 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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