SATHEE: Redox Reactions And Electrochemistry Ques 220

Redox Reactions And Electrochemistry Ques 220

Question: The e.m.f. of the cell $Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e$ at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]

Options:

A) $\frac{0.32}{e^{0.0295}}$

B) $\frac{0.32}{10^{0.0295}}$

C) $\frac{0.26}{10^{0.0295}}$

D) $\frac{0.32}{10^{0.0591}}$

Show Answer

Answer:

Correct Answer: B

Solution:

For this cell, reaction is: $Z n + F e^{2 +} \to Z n^{2 +} + F e$ $E = E^{0} - \frac{0.0591}{n} \log \frac{c_{1}}{c_{2}}$ ;

$E^{0} = E + \frac{0.0591}{n} \log \frac{c_{1}}{c_{2}}$ $= 0.2905 + \frac{0.0591}{2} \log \frac{10^{- 2}}{10^{- 3}} = 0.32 V$ .

$E^{0} = \frac{0.0591}{2} \log K_{c}$ ; $\log K_{c} = \frac{0.32 \times 2}{0.0591} = \frac{0.32}{0.0295}$
$∴ K_{c} = \frac{0.32}{10^{0295}}$ .

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Redox Reactions And Electrochemistry Ques 220

Question: The e.m.f. of the cell Zn|Zn2+(0.01M)||Fe2+(0.001M)|Fe at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]

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Question: The e.m.f. of the cell $Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e$ at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]