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Electrochemistry question

2021 · 31 Aug · Shift 1 · Q14
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Electrochemistry question

2021 · 31 Aug · Shift 1 · Q14

JEE MainChemistryElectrochemistryNumerical+4 / −1
Consider the following cell reaction : Cd(s)+Hg2SO4(s)+95H2O(l)⇌C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \over 5}{H_2}{O_{(l)}}\rightleftharpoonsCd(s)​+Hg2​SO4(s)​+59​H2​O(l)​⇌ CdSO4.95H2O(s)+2Hg(l)CdS{O_4}.{9 \over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}CdSO4​.59​H2​O(s)​+2Hg(l)​ The value of Ecell0E_{cell}^0Ecell0​ is 4.315 V at 25 ∘^\circ∘ C. If Δ\DeltaΔ H ∘^\circ∘=−-− 825.2 kJ mol −-− 1, the standard entropy change Δ\DeltaΔ S ∘^\circ∘ in J K −-− 1 is ‾\underline{\hspace{2cm}}​. (Nearest integer) [Given : Faraday constant = 96487 C mol −-− 1]
Numerical answer
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