E- for the electrochemical cell- Zns - Zn2- 1 Maq- - Cu2- 1 M aq- - Cus is 1-10 V at 25-C- The equilibrium constant for the cell reaction- Zns - Cu2- aq- Zn2-aq- - Cus will be-

The correct option is D 10^37 Given Data : 1) Electrode Potential (E^ #176;) = 1.10V #160; #160; #160; #160; #16 ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

E∘ for the el...

Question

$E^{\circ}$ for the electrochemical cell, $Z n (s) | Z n^{2 +} 1 M (a q .) ∥ C u^{2 +} 1 M (a q .) | C u (s)$ is $1.10 V$ at $25^{\circ} C$ . The equilibrium constant for the cell reaction, $Z n (s) + C u^{2 +} (a q .) ⇌ Z n^{2 +} (a q .) + C u (s)$ will be:

10^{- 38}

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10^{37}

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10^{- 39}

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10^{39}

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Solution

The correct option is D $10^{37}$
$G i v e n$ $D a t a$ : 1) $E l e c t r o d e P o t e n t i a l$ $(E^{\circ})$ = $1.10 V$
2) $T e m p e r a t u r e$ $(T)$ = $25^{\circ} C$ = $400 K$
3) $n$ = $n o$ $o f$ $m o l e s$ $o f$ $e l e c t r o l y t e$ = $2$

$F o r$ $e q u i l i b r i u m$ $c o n s t a n t$ $b y$ $N e r n s t^{'} s$ $e q u a t i o n$ $w e$ $h a v e$ ,

$K$ = $a n t i l o g$ $[\frac{n (E^{\circ})}{0.0591}]$

= $a n t i l o g$ $[\frac{2 \times 1.10}{0.0591}]$

= $1.67$ $\times 10$ $^{37}$

Hence the correct option is 'B'.

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Similar questions

E^{\circ}

for the electrochemical cell,

Z n (s) | Z n^{2 +} (1 M) a q . | C u (s)

1.10 v o l t

25^{\circ} C

.
The equilibrium constant for the cell reaction:

Z n (s) + C u^{2} (a q .) ⇌ C u (s) + Z n^{2 +} (a q .)

will be:

$E^{0}$ for the cell, $Z n | Z n {^{2 +}}_{(a q)} | | C u {^{2 +}}_{(a q)} | C u$ is $1.10 V$ at $25^{0} C$ . The equilibrium constant for $Z n + C u {^{2 +}}_{(a q)} \to Z n {^{2 +}}_{(a q)} + C u$ the cell reaction is of the order of :

E^{\circ}

for the cell,

{Zn(s)|Zn}^{2 +} (aq) | | {Cu}^{2 +} (aq)|Cu(s)

1.10

V at

25^{\circ} C

. The equilibrium constant for the reaction,

Zn(s) + {Cu}^{2 +} (aq) \to Cu(s) + {Zn}^{2 +} (aq)

1.94 \times 10^{x}

. The value of

x

E^{\circ}

for the cell is

Z n | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u

25^{\circ} C

, the equilibrium constant for the reaction

Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)

is of the order of

E^{\circ}

for the cell is

Z n | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u

25^{\circ} C

, the equilibrium constant for the reaction

Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)

is of the order of

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E∘ for the electrochemical cell, Zn(s)|Zn2+1 M(aq.)∥Cu2+1 M(aq.)|Cu(s) is 1.10 V at 25∘C. The equilibrium constant for the cell reaction, Zn(s)+Cu2+(aq.)⇌Zn2+(aq.)+Cu(s) will be:

The correct option is D 1037GivenData: 1)ElectrodePotential (E∘)= 1.10V 2) Temperature (T)=25∘C=400K 3) n =no of moles of electrolyte=2For equilibrium constant by Nernst′s equation we have,K= antilog [n(E∘)0.0591] = antilog [2×1.100.0591] = 1.67 ×10 37Hence the correct option is 'B'.

$E^{\circ}$ for the electrochemical cell, $Z n (s) | Z n^{2 +} 1 M (a q .) ∥ C u^{2 +} 1 M (a q .) | C u (s)$ is $1.10 V$ at $25^{\circ} C$ . The equilibrium constant for the cell reaction, $Z n (s) + C u^{2 +} (a q .) ⇌ Z n^{2 +} (a q .) + C u (s)$ will be: