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JEE CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Analyzing +4 | -1
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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Correct option: A
142 mV142 \mathrm{~mV}
Option A (142 mV) is correct because it involves calculating the cell potential using the Nernst equation, considering the pH and solubility product.

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