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Equivalent, Molar Conductivity and Cell Constant

For a reactio...

Question

For a reaction ${Z n}^{+ 2} + {2 e}^{-} ⟶ Z n, E^{o} = - 0.76 V$ which of the following statement is true? (E=Electrode potential)

E depends upon temperature

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E changes when Zn plate is made bigger in size

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E becomes double if equation is multiplied by2

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{Z n}^{+ 2}

can be reduced to Zn by

H_{2}

gas

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Solution

The correct option is A E depends upon temperature
$Δ$ G $= - n F E$ . $Δ$ G varies with Temperature and hence E also varies with Temperature.
E doesnt change when plate is made bigger size.
E doesn't change if Eq is multiplied by 2
${Z n}^{+ 2}$ can't be reduced to Zn with $H_{2}$
So Option A is True

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Similar questions

E^{o}

Z n ⟶ {Z n}^{2 +} + 2 e^{-}

E^{o} = + 0.76 V

F e ⟶ {F e}^{2 +} + 2 e^{-}

E^{o} = + 0.41 V

{F e}^{2 +} + Z n ⟶ {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^{o}

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

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

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

E^{o}

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

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

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

{Z n}^{2 +} / Z n, E^{o} = - 0.76 V, f o r {A g}^{+} / A g, E^{o} = 0.799 V

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