MCQ
Cell equation: $A +2 B ^{-} \longrightarrow A ^{2+}+2 B $
$A ^{2+}+2 e ^{-} \longrightarrow A$ $E^0= +0.34$ V and $log_{10}$ k = 15.6 at 300 K for cell reactions find $E^0$ for $B ^{+}+ e ^{-} \longrightarrow B$
  • 0.80
  • B
    1.26
  • C
    – 0.54
  • D
    – 10.94

Answer

Correct option: A.
0.80
0.80
Explanation:
$A +2 B ^{-} \longrightarrow A ^{2+}+2 B$
Half reaction anode: $A \longrightarrow A ^{2+}+2 e ^{-}$
`"E"_"ox"^0` = –0.34 V ........[Given: $A ^{2+}+2 e ^{-} \longrightarrow A$ = +0.34 V
Cathode: $2 B ^{+}+2 e ^{-} \longrightarrow 2 B$ `"E"_"red"^0` = ?
$log_{10}$ K = 156; T = 300 K; n = 2;
F = 96500 C
$\mathrm{R = 8.314 JK^{-1} mol^{-1}}$
$∆G^\circ= – 2.303$ RT log K
$∴ nFE^\circ= – 2.303$ 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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