Find the E0 value for the reaction-In2-aq-Cu2-aq- In3-aq-Cu-aq at 298 K-Given-E0Cu2-aq-Cu-aq- 0-15 VE0 In2-aq-In-aq- -0-4 VE0 In3-aq-In-aq- -0-42 V

Δ G^0 = nFE^0 The given reversible reaction can be obtained as, Cu^2+(aq)+e^ → Cu^+(aq)......(1); nbsp; Δ G^0_1 nbsp;= 0.15F nbsp; In^2+(aq)+e^ → In^+ ...

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Byju's Answer

Standard XII

Chemistry

Order of Reaction

Find the E0 v...

Question

Find the $E^{0}$ value for the reaction,
$I n^{2 +} (a q) + C u^{2 +} (a q) ⇌ I n^{3 +} (a q) + C u^{+} (a q)$ at $298 K$ .
Given:
$E_{C u^{2 +} (a q) / C u^{+} (a q)}^{0} = 0.15 V E_{I n^{2 +} (a q) / I n^{+} (a q)}^{0} = - 0.4 V$
$E_{I n^{3 +} (a q) / I n^{+} (a q)}^{0} = - 0.42 V$

0.59 V

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0.97 V

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1.23 V

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0.23 V

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Solution

The correct option is A $0.59 V$
$Δ G^{0} = - n F E^{0}$
The given reversible reaction can be obtained as,
$C u^{2 +} (a q) + e^{-} \to C u^{+} (a q) . . . . . . (1); Δ G_{1}^{0} = - 0.15 F$
$I n^{2 +} (a q) + e^{-} \to I n^{+} (a q) . . . . . . (2); Δ G_{2}^{0} = + 0.40 F$
$I n^{3 +} (a q) + 2 e^{-} \to I n^{+} (a q) . . . . . . . (3); Δ G_{3}^{0} = + 0.84 F$
Reversing equation (3), we get,
$I n^{+} (a q) \to I n^{3 +} (a q) + 2 e^{-} . . . . . . (4); Δ G_{4}^{0} = - 0.84 F$
Adding equation (1), (2) and (4), we get,
$C u^{2 +} (a q) + I n^{2 +} (a q) ⇌ I n^{3 +} (a q) + C u^{+} (a q); Δ G^{0} = - F E^{o}$
$∵$
Number of electron involved in the reaction is 1.
$Δ G^{0} = Δ G_{1}^{0} + Δ G_{2}^{0} + Δ G_{4}^{0}$
$Δ G^{0} = - 0.59 F$
$- F E^{0} = - 0.59 F$
$∴$
$E^{0} = 0.59 V$

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Similar questions

Q. Find the

E^{0}

value for the reaction,

I n^{2 +} (a q) + C u^{2 +} (a q) ⇌ I n^{3 +} (a q) + C u^{+} (a q)

298 K

.
Given:

E_{C u^{2 +} (a q) / C u^{+} (a q)}^{0} = 0.15 V E_{I n^{2 +} (a q) / I n^{+} (a q)}^{0} = - 0.4 V

E_{I n^{3 +} (a q) / I n^{+} (a q)}^{0} = - 0.42 V

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C u^{2 +} (a q) + e^{-} \to C u^{+} (a q) and C u^{2 +} (a q) + e^{-} \to C u (s) are + 0.15 V and + 0.50 V

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