Question

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

Answer 1

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