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JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Analyzing +4 | -1
Two platinum electrodes were immersed in a solution of cupric sulphate and electric current passed through the solution. After sometime, it was found that the colour of copper sulphate disappeared with evolution of gas at the electrode. The colourless solution contains:
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Correct option: D
Sulphuric acid
Option D (Sulphuric acid) is correct because during electrolysis of CuSO4CuSO_4, copper is deposited at the cathode and oxygen is evolved at the anode, leaving sulphuric acid in the solution.

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)+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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