E- for the cell- Zns-Zn2-aq-Cu2-aq-Cus is 1-10 V at 25-C- The equilibrium constant for the reaction- Zns-Cu2-aq-Cus-Zn2-aq is 1-94 - 10 x- The value of x is

According to Nernst equation E_cell=E^∘_cell 0.059n log[Zn^2+][Cu^2+] At equilibrium, E_cell= 0 Given, E^∘_cell=1.10 V ∴ 0=1.1 0.0592 log K_eq ∴ K_eq=1.94× 10^3 ...

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Standard VII

Chemistry

Equilibrium Constant from Nernst Equation

E∘ for the ce...

Question

$E^{\circ}$ for the cell,
${Zn(s)|Zn}^{2 +} (aq) | | {Cu}^{2 +} (aq)|Cu(s)$ is $1.10$ V at $25^{\circ} C$ . The equilibrium constant for the reaction, $Zn(s) + {Cu}^{2 +} (aq) \to Cu(s) + {Zn}^{2 +} (aq)$ is $1.94 \times 10^{x}$ . The value of $x$ is

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Solution

The correct option is B $37$
According to Nernst equation
$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} log \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
At equilibrium, $E_{cell}$ = $0$
Given, $E_{cell}^{\circ} = 1.10 V$
$∴ 0 = 1.1 - \frac{0.059}{2} log K_{e q}$
$∴ K_{e q} = 1.94 \times 10^{37}$

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$E^{0}$ 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 :

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E∘ for the cell, Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) is 1.10 V at 25∘C. The equilibrium constant for the reaction, Zn(s)+Cu2+(aq)→Cu(s)+Zn2+(aq) is 1.94×10x. The value of x is

The correct option is B 37According to Nernst equation Ecell=E∘cell−0.059n log[Zn2+][Cu2+] At equilibrium, Ecell= 0 Given, E∘cell=1.10 V ∴0=1.1−0.0592 log Keq ∴Keq=1.94×1037

$E^{\circ}$ for the cell,
${Zn(s)|Zn}^{2 +} (aq) | | {Cu}^{2 +} (aq)|Cu(s)$ is $1.10$ V at $25^{\circ} C$ . The equilibrium constant for the reaction, $Zn(s) + {Cu}^{2 +} (aq) \to Cu(s) + {Zn}^{2 +} (aq)$ is $1.94 \times 10^{x}$ . The value of $x$ is

The correct option is B $37$
According to Nernst equation
$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} log \frac{[Z n^{2 +}]}{[C u^{2 +}]}$
At equilibrium, $E_{cell}$ = $0$
Given, $E_{cell}^{\circ} = 1.10 V$
$∴ 0 = 1.1 - \frac{0.059}{2} log K_{e q}$
$∴ K_{e q} = 1.94 \times 10^{37}$