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प्रश्न
Cell equation: \[\ce{A + 2B^- -> A^{2+} + 2B}\]
\[\ce{A^{2+} + 2e^- -> A}\] E0 = +0.34 V and log10 k = 15.6 at 300 K for cell reactions find E0 for \[\ce{B^+ + e^- -> B}\]
पर्याय
0.80
1.26
– 0.54
– 10.94
उत्तर
0.80
Explanation:
\[\ce{A + 2B^- -> A^{2+} + 2B}\]
Half reaction anode: \[\ce{A -> A^2+ + 2e^-}\]
`"E"_"ox"^0` = –0.34 V ........[Given: \[\ce{A^{2+} + 2e^- -> A}\] E0 = +0.34 V]
Cathode: \[\ce{2B^+ + 2e^- -> 2B}\] `"E"_"red"^0` = ?
log10 K = 156; T = 300 K; n = 2;
F = 96500 C
R = 8.314 JK−1 mol−1
∆G° = – 2.303 RT log K
∴ nFE° = – 2.303 RT log K
`"E"_"cell"^0 = (2.303 "RT" log "K")/("nF")`
= `(2.303 xx 8.314 xx 300 xx 15.6)/(2 xx 96500)`
= 0.4643 V
`"E"_"cell"^0 = "E"_"ox"^0 + "E"_"red"^0`
`"E"_"red"^0 = "E"_"cell"^0 + "E"_"ox"^0`
∴ 0.4643 – (–0.34)
= 0.4643 + 0.34
= 0.8043
= 0.80 V
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