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Factors Affecting Electrolysis

What does neg...

Question

What does negative sign in the expression $E^{0}_{Z n^{2 +} / Z n} = - 0.76 V$ mean?

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Solution

When zinc electrode is joined with SHE, the Zinc electrode will act as anode (negative electrode) i.e., oxidation occurs on this electrode spontaneously.

The negative sign of standard reduction potential indicates that the electrode when joined with SHE acts as an anode and oxidation occurs on this electrode spontaneously.

Thus, $E^{0}_{Z n^{2 +} / Z n} = - 0.76 V$ signifies spontaneous oxidation of Zn at anode.

Corresponding reaction at anode is:
$Z n \to Z n^{2 +} + 2 e^{-}$

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Similar questions

$Z n | Z n^{2 +} (C_{1}) | | Z n^{2 +} (C_{2}) | Z n$ ,

for this cell $Δ G$ is negative if:

Z n | {Z n}^{2 +} (C_{1}) ∥ {Z n}^{2 +} (C_{2}) | Z n

Δ G

is negative if :-

Q. Which of the following statement(s) is/are true?

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

Q. Calculate the equilibrium constant for the reaction
Cd2+(aq) + Zn(s) ----> Zn2+(aq) + Cd(s)
If E0 Cd2+/Cd = -0.403 V
E0 Zn2+/Zn = -0.763 V

Q. Consider the follwoing reduction processes :

Z n^{2 +} + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C a^{2 +} + 2 e^{-} \to C a (s); E^{0} = - 2.87 V

M g^{2 +} + 2 e^{-} \to M g (s); E^{0} = - 2.36 V

N i^{2 +} + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

The reducing power of the metals increases in the order :

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Factors Affecting Electrolysis

Standard X Chemistry

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