At 298 K- the equilibrium constant for the reaction Zn2- - 4 NH3 - -ZnNH34-2 is 10oIf Eo-ZnNH34-2- -Zn- 4NH3- - -1-03 V then the value of EoZn-Zn2- will be-

The correct option is C 0.7654 V Zn ^ 2+ +4 NH _ 3 ⇌[ Zn( NH _ 3 ) #160; _ 4 ] #160; ^ 2+ We know, log K _ eq = n 0.059 × E ^ 0 n=2 ...

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EMF

At 298 K, t...

Question

At $298 K$ , the equilibrium constant for the reaction
$Z n^{2 +} + 4 N H_{3} ⇌ [Z n (N H_{3})_{4}]^{2}$ is $10^{o}$
If $E_{[Z n (N H_{3})^{4}]^{2} / [Z n, 4 N H_{3}]}^{o} = - 1.03 V$ then the value of $E_{Z n / Z n^{2 +}}^{o}$ will be:

- 0.7654 V

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- 1.1 V

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+ 1.1 V

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none of these

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Solution

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Similar questions

298

Z n^{2 +} + 4 N H_{3} ⇌ [Z n (N H_{3})_{4}]^{2 +}

10^{9}

E_{[Z n (N H_{3})_{4}]^{2 +} / [Z n^{2 +} + 4 N H_{3}]}^{o} = - 1.03

E_{Z n / Z n^{2 +}}^{o}

298 K

{Z n}^{+ 2} + 4 {N H}_{3} ⇌ {[Z n {({N H}_{3})}_{4}]}^{+ 2}

10^{9}

E_{{[Z n {({N H}_{3})}_{4}]}^{+ 2} / Z n + 4 {N H}_{3}}^{o} = - 1.03 V

E_{Z n / {Z n}^{+ 2}}

Z n^{2 +}

{[Z n {(N H_{3})}_{4}]}^{2 +}

{Z n^{2 +}}_{(a q)} + 4 N H_{3} ⇌ {[Z n {(N H_{3})}_{4}]}^{2 +}_{(a q)}

Z n^{2 +}

Z n^{2 +} (a q) + 4 N H_{3} (a q) ⇌ [Z n (N H_{3})_{4}]^{2 +} (a q)

The EMF 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 constant for the cell reaction is :

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