E _ 1 , E _ 2 and E _ 3 are the emf of the following three galvanic cells respectively (i) Zn (s)| Zn ^ 2 + (0 . 1M)|| Cu ^ 2 + (1M)| Cu( s) (ii) Zn (s) | Zn ^ 2 + (1M)|| Cu ^ 2 + (1M)|Cu(s) (iii) Zn (s)| Zn ^ 2 + (1M)|

Correct answer: B. E _ 1 > E _ 2 > E _ 3 Key-idea Use Nernst equation, i.e. E(emf) = E ° _ cell - (0 . 0591)/(2)log [ Zn ^ 2 + ] [ Cu ^ 2 + ] For reaction (i) E _ 1 = E ° _ (cell) - (0 . 0591)/(2)log(0 . 1)/(1) E _ 1 = E ° _ (cell) + (0 . 0591)/(2) For reaction (ii) E _ 2 = E ° _ (cell) - (0 . 0591)/(2)log(1)/(1) = 0 E _ 2 = E ° _ (cell) For reaction (iii) E _ 3 = E ° _ (cell) - (0 . 0591)/(2)log(1)/(0 . 1) E _ 3 = E ° _ (cell) - (0 . 0591)/(2) Hence, E _ 1 > E _ 2 > E _ 3 .

BITSAT 2019

Exam: BITSAT
Year: 2019
Subject: Chemistry
Chapter: Electrochemistry

BITSAT 2019 · Chemistry · Electrochemistry

E _ 1 , E _ 2 and E _ 3 are the emf of the following three galvanic...

${E}_{1}, {E}_{2}$ and ${E}_{3}$ are the emf of the following three galvanic cells respectively

(i)

Zn \left(s\right)\left|{Zn}^{2 +} \left(0 . 1M\right)\right|\left|{Cu}^{2 +}\left(1M\right)\right|

Cu\left(

s\right)

(ii) $Zn \left(s\right)$ \left|{Zn}^{2 +} \left(1M\right)\right|\left|{Cu}^{2 +}\left(1M\right)\right|Cu\left(s\right) $(iii)$ Zn \left(s\right)\left|{Zn}^{2 +} \left(1M\right)\right| \left|{Cu}^{2 +}\left(0 . 1M\right)\right| $Cu\left($ s\right) $Which of the following is true?$

Options

A ${E}_{2} > {E}_{1} > {E}_{3}$
B ${E}_{1} > {E}_{2} > {E}_{3}$
C ${E}_{3} > {E}_{1} > {E}_{2}$
D ${E}_{3} > {E}_{2} > {E}_{1}$

Correct answer

B. ${E}_{1} > {E}_{2} > {E}_{3}$

Step-by-step solution

Key-idea

Use Nernst equation, i.e.

$E\left(emf\right) = E{°}_{cell} - \dfrac{0 . 0591}{2}log\dfrac{\left[{Zn}^{2 +}\right]}{\left[{Cu}^{2 +}\right]}$

For reaction \left(i\right) $$ {E}_{1} = E{°}_{\left(cell\right)} - \dfrac{0 . 0591}{2}log\dfrac{0 . 1}{1} $$ {E}_{1} = E{°}_{\left(cell\right)} + \dfrac{0 . 0591}{2} $For reaction \left(ii\right)$

${E}_{2} = E{°}_{\left(cell\right)} - \dfrac{0 . 0591}{2}log\dfrac{1}{1} = 0$

${E}_{2} = E{°}_{\left(cell\right)}$

For reaction \left(iii\right) $$ {E}_{3} = E{°}_{\left(cell\right)} - \dfrac{0 . 0591}{2}log\dfrac{1}{0 . 1} $$ {E}_{3} = E{°}_{\left(cell\right)} - \dfrac{0 . 0591}{2} $Hence,$ {E}_{1} > {E}_{2} > {E}_{3}.

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