SATHEE: Redox Reactions And Electrochemistry Ques 339

Redox Reactions And Electrochemistry Ques 339

Question: For $A g \to A g^{+} + e^{-},$ $E^{\circ} = - 0.798 V$ $V^{2 +} + V O^{2 +} + 2 H^{+} \to 2 V^{3 +} + H_{2} O,$ $E^{\circ} = - 0.614 V$ $V^{3 +} + A g^{+} + H_{2} O \to V O^{2 +} + 2 H^{+} + A g,$ $E^{\circ} = -, 0.438 V$ Then $E^{\circ}$ for the reaction $V^{3 +} + e^{-} \to V^{2 +}$ is

Options:

A) + 0.255V

B) $-, 0.255, V$

C) $-, 0.254, V$

D) $-, 1.055, V$

Show Answer

Answer:

Correct Answer: C

Solution:

[b] $A g \to A g^{+} + e^{-},$ $E^{o} = -, 0.798 V$ $V^{3} + A g^{+} + H_{2} O \to V O^{2 +} + 2 H^{+} + A g,$ $E^{o} = - 0.438 V$ Adding, we get
$∴ V^{3 +} + H_{2} O \to V O^{2 +} + 2 H^{+} + e^{-,}$

$E^{o} = - 0.36 V$ $V^{2 +} + V O^{2 +} + 2 H^{+} \to 2 V^{3 +} + H_{2} O,$ $E^{o} = - 0.614 V$ Adding $V^{2 +} \to V^{3 +} + e^{-}$ $E^{o} = - 0.254 V$ $V^{3 +} + e^{-} \to V^{2 +}$ $E^{o} = - 0.254 V$

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Redox Reactions And Electrochemistry Ques 339

Question: For Ag→Ag++e−, E∘=−0.798V V2++VO2++2H+→2V3++H2O, E∘=−0.614V V3++Ag++H2O→VO2++2H++Ag, E∘=−,0.438V Then E∘ for the reaction V3++e−→V2+ is

Options:

Answer:

Solution:

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Question: For $A g \to A g^{+} + e^{-},$ $E^{\circ} = - 0.798 V$ $V^{2 +} + V O^{2 +} + 2 H^{+} \to 2 V^{3 +} + H_{2} O,$ $E^{\circ} = - 0.614 V$ $V^{3 +} + A g^{+} + H_{2} O \to V O^{2 +} + 2 H^{+} + A g,$ $E^{\circ} = -, 0.438 V$ Then $E^{\circ}$ for the reaction $V^{3 +} + e^{-} \to V^{2 +}$ is