SATHEE: Redox Reactions And Electrochemistry Ques 609

Redox Reactions And Electrochemistry Ques 609

Question: The electrode potential $E_{(Z n^{2 +} / Z n)}$ of a zinc electrode at $25^{\circ} C$ with an aqueous solution of $0.1 M Z n S O_{4}$ is $[E {^{\circ}}_{(Z n^{2 +} / Z n)} = - 0.76 V$ . Assume $\frac{2.303 R T}{F} = 0.06 a t 298 K] .$

Options:

A) +0.73

B) $- 0.79$

C) $- 0.82$

D) $- 0.70$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] For $Z n^{2 +} \overset{}{\to} Z n$ $E_{Z n^{2 +} / Z n} = E {^{\circ}}_{Z n^{2 +} / Z n} - \frac{2.303 R T}{n F} \log \frac{[Z n]}{[Z n^{2 +}]}$

$= - 0.76 - \frac{0.06}{2} l o g \frac{1}{[0.1]} = - 0.76 - 0.03$ $E_{Z n^{2 +} / Z n} = - 0.79 V$

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Redox Reactions And Electrochemistry Ques 609

Question: The electrode potential E(Zn2+/Zn) of a zinc electrode at 25∘C with an aqueous solution of 0.1MZnSO4 is [E∘(Zn2+/Zn) =−0.76V . Assume 2.303RTF=0.06at298K].

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Question: The electrode potential $E_{(Z n^{2 +} / Z n)}$ of a zinc electrode at $25^{\circ} C$ with an aqueous solution of $0.1 M Z n S O_{4}$ is $[E {^{\circ}}_{(Z n^{2 +} / Z n)} = - 0.76 V$ . Assume $\frac{2.303 R T}{F} = 0.06 a t 298 K] .$