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JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Medium
1 of 1
Analyzing +4 | -1
Statement I: A metal having negative reduction
potential when dipped in the solution of its own
ions has a tendency to pass into the solution.

Statement II: Metals having negative reduction
potential have large hydration energy.
View answer and explanation
Correct option: B
If both statements are CORRECT, and Statement
II is NOT the CORRECT explanation of
Statement I
Option B is correct because a metal with a negative reduction potential tends to pass into solution, and metals with negative reduction potential have large hydration energy, but the hydration energy is not the direct explanation for the tendency to pass into solution.

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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}

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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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