Calculate the half-cell potential at 298 K for the reaction- Zn2- - 2e- Znif -Zn2- - 0-1 M and E- - -0-76 volt-

The correct option is B 0.789 volt .According to Nerast Equation E_m^n+/m = E^∘m^n+/m 0.0591n log 1[m^n+] E_Zn^2+/zn = E_zn^2+/zn 0.05912 log ...

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Question

Calculate the half-cell potential at $298 K$ for the reaction,
$Z n^{2 +} + 2 e^{-} \to Z n$
if $[Z n^{2 +}] = 0.1 M$ and $E^{\circ} = - 0.76 v o l t$ .

- 0.847 v o l t

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- 0.789 v o l t

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- 0.475 v o l t

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- 0.366 v o l t

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Similar questions

E^{\circ}

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = + 0.76 v o l t

F e \to F e^{2 +} + 2 e^{-}; E^{\circ} = + 0.41 v o l t

F e^{2 +} + Z n \to Z n^{2 +} + F e

Z n + 2 H^{+} (a q .) \to Z n^{2 +} (0.1 M) + H_{2} (g) 1 a t m

0.28

25^{\circ} C

p H

E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 v o l t

E_{H^{+} / H_{2}}^{\circ} = 0

E^{⊝}

Z n \to Z n^{2 +} + 2 e^{-}; E^{⊝} = + 0.76 V

F e \to F e^{2 +} + 2 e^{-}; E^{⊝} = 0.41 V

F e^{2 +} Z n \to Z n^{2 +} + F e

Z n \to Z n^{2 +} + 2 e^{-} (E_{(Z n^{2 +} / Z n)}^{0} = - 0.76 V)

F e \to F e^{2 +} + 2 e^{-} E^{0} = 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

E^{\circ}

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = 0.76 V

C u \to C u^{2 +} + 2 e^{-}; E^{\circ} = 0.34 V

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

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