The standard reduction potential data at 25∘C is given below.E∘(Fe3+/Fe2+)=+0.77 V;E∘(Fe2+/Fe)=−0.44 VE∘(Cu2+/Cu)=+0.34 V;E∘(Cu+/Cu)=+0.52 VE∘(O2(g)+4H++4e−)⟶2H2O)=+1.23 VE∘(O2(g)+2H2O+4e−)⟶4OH−)=+0.40 VE∘(Cr3+/Cr)=−0.74 V;E∘(Cr2+/Cr)=+0.91 VMatch E∘ of the rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.

ΔGnet ∘−n3FE∘3∴E3∘=ΔG∘1+ΔG∘=−n1FE∘1−n2FE∘2=n3n1E1∘+n2E2∘(P) E3∘Fe3+/FeNet reaction Fe3+⟶FeE3∘=n3n1E1∘+n2E2∘=30.77+2(−0.44)=3−0.11=−0.04 VThus, P - (3)Net reaction4H2O⇌4H++4OH−E3∘=n3n1E1∘+n2E2∘=E1∘+E2∘=−1.23+0.40=−0.83 VThus, Q−(4)(R) Cu2++Cu⟶2Cu+For thus E∘ of Cu2+⟶Cu+E3∘=n3n1E1∘+n2E2∘=12×0.34+1×(−0.52)=0.16 VE∘=−0.52+0.16=−0.36 VThus, (R) - (1)(S) Cr3+⟶Cr2+E3∘=1−0.74×3+2×0.91=−0.4 VThus, S=(2)P−(3),Q−(4),R−(1),S−(2)

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Standard reduction potential

The standard reduction potential data at 25^{} {C} is given be

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The standard reduction potential data at $2 5^{\circ} C$ is given below.
$E^{\circ} (Fe^{3 +} / Fe^{2 +}) = + 0.77 V; E^{\circ} (Fe^{2 +} / Fe) = - 0.44 V E^{\circ} (Cu^{2 +} / Cu) = + 0.34 V; E^{\circ} (Cu^{+} / Cu) = + 0.52 V E^{\circ} (O_{2} (g) + 4 H^{+} + 4 e^{-}) ⟶ 2 H_{2} O) = + 1.23 V E^{\circ} (O_{2} (g) + 2 H_{2} O + 4 e^{-}) ⟶ 4 O H^{-}) = + 0.40 V E^{\circ} (Cr^{3 +} / Cr) = - 0.74 V; E^{\circ} (Cr^{2 +} / Cr) = + 0.91 V$
Match $E^{\circ}$ of the rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.

P 4 Q 1 R 2 S 3

P 2 Q 3 R 4 S I

P 1 Q 2 R 3 S 4

P 3 Q 4 R 1 S 2

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JEE Advanced 2018

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Chloe Discussed

The standard reduction potential data at

2 5^{\circ} C

is given below.

E^{\circ} (Fe^{3 +} / Fe^{2 +}) = + 0.77 V; E^{\circ} (Fe^{2 +} / Fe) = - 0.44 V E^{\circ} (Cu^{2 +} / Cu) = + 0.34 V; E^{\circ} (Cu^{+} / Cu) = + 0.52 V E^{\circ} (O_{2} (g) + 4 H^{+} + 4 e^{-}) ⟶ 2 H_{2} O) = + 1.23 V E^{\circ} (O_{2} (g) + 2 H_{2} O + 4 e^{-}) ⟶ 4 O H^{-}) = + 0.40 V E^{\circ} (Cr^{3 +} / Cr) = - 0.74 V; E^{\circ} (Cr^{2 +} / Cr) = + 0.91 V

Match

E^{\circ}

of the rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.

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Text solutionVerified

Δ G_{net}^{\circ} - n_{3} F E^{\circ}_{3} ∴ E_{3}^{\circ} = Δ G^{\circ}_{1} + Δ G^{\circ} = - n_{1} F E^{\circ}_{1} - n_{2} F E^{\circ}_{2} = \frac{n _{1} E _{1}^{\circ} + n _{2} E _{2}^{\circ}}{n _{3}}

(P)

E_{3}^{\circ} Fe^{3 +} / Fe

Net reaction

Fe^{3 +} ⟶ Fe

E_{3}^{\circ} = \frac{n _{1} E _{1}^{\circ} + n _{2} E _{2}^{\circ}}{n _{3}} = \frac{0.77 + 2 ( - 0.44 )}{3} = \frac{- 0.11}{3} = - 0.04 V

Thus,

P

- (3)
Net reaction

4 H_{2} O ⇌ 4 H^{+} + 4 OH^{-}

E_{3}^{\circ} = \frac{n _{1} E _{1}^{\circ} + n _{2} E _{2}^{\circ}}{n _{3}} = E_{1}^{\circ} + E_{2}^{\circ} = - 1.23 + 0.40 = - 0.83 V

Thus,

Q - (4)

(R)

Cu^{2 +} + Cu ⟶ 2 Cu^{+}

For thus

E^{\circ}

Cu^{2 +} ⟶ Cu^{+}

E_{3}^{\circ} = \frac{n _{1} E _{1}^{\circ} + n _{2} E _{2}^{\circ}}{n _{3}} = \frac{2 \times 0.34 + 1 \times ( - 0.52 )}{1} = 0.16 V

E^{\circ} = - 0.52 + 0.16 = - 0.36 V

Thus, (R) - (1)
(S)

Cr^{3 +} ⟶ Cr^{2 +}

E_{3}^{\circ} = \frac{- 0.74 \times 3 + 2 \times 0.91}{1} = - 0.4 V

Thus,

S = (2)

P - (3), Q - (4), R - (1), S - (2)

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Question Text	The standard reduction potential data at $2 5^{\circ} C$ is given below. $E^{\circ} (Fe^{3 +} / Fe^{2 +}) = + 0.77 V; E^{\circ} (Fe^{2 +} / Fe) = - 0.44 V E^{\circ} (Cu^{2 +} / Cu) = + 0.34 V; E^{\circ} (Cu^{+} / Cu) = + 0.52 V E^{\circ} (O_{2} (g) + 4 H^{+} + 4 e^{-}) ⟶ 2 H_{2} O) = + 1.23 V E^{\circ} (O_{2} (g) + 2 H_{2} O + 4 e^{-}) ⟶ 4 O H^{-}) = + 0.40 V E^{\circ} (Cr^{3 +} / Cr) = - 0.74 V; E^{\circ} (Cr^{2 +} / Cr) = + 0.91 V$ Match $E^{\circ}$ of the rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.
Topic	Electrochemistry
Subject	Chemistry
Class	Class 12
Answer Type	Text solution:1
Upvotes	71

Text solutionVerified

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