Which of the following will act as cathode when connected to standard hydrogen electrode which has $E^{0}$ value given as zero?
i) $Z n / Z n + 2, E^{0} = - 0.76 V$
ii) $C u^{+ 2} / C u, E^{0} = + 0.34 V$
iii) $A l / A l^{+ 3}, E^{0} = - 1.66 V$
iv) $H g^{2 +} / H g, E^{0} = + 0.885 V$

i and ii

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ii and iv

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i and iii

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i,ii,iii and iv

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Solution

The correct option is B ii and iv
In this standard hydrogen electrode, the hydrogen electrode is acting as anode. Therefore, in hydrogen or anode half cell oxidation will take place. Reaction occurring at anode,
$H_{2} ⇌ 2 H^{+} + 2 e^{-}$ , $E^{0} = 0$
At cathode reduction should take place. Metals lying below hydrogen in the reactivity series have positive value of reduction potential and are poor reducing agents and good oxidising agents. Thus, they reduce themselves. Among the given metals $C u$ and $H g$ lie below $H$ and have positive reduction potentials. Hence, they can act as cathode.
Reaction occurring at cathode,
$C u^{+ 2} + 2 e^{-} ⇌ C u (s)$ ; $E^{0} = + 0.34 V$
$H g^{+ 2} + 2 e^{-} ⇌ H g (s)$ ; $E^{0} = + 0.885 V$

Suggest Corrections

Similar questions

Z n \to Z n^{+ +} + 2 e^{-}, E^{0} = 0.76 V

C u \to C u^{+ +} + 2 e^{-}, E^{0} = 0.34 V

M g^{2 +} + 2 e^{-} \to M g; E^{0} = - 2.37 V

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

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.44 V

E^{0}

Z n^{2 +} / Z n = - 0.76 V

A g^{+} / A g = + 0.08 V

C u^{2 +} / C u = 0.34 V

C u^{2 +} + 2 e^{-} ⟶ C u

E^{0} = 0.337 V

C u^{2 +} + e^{-} ⟶ C u^{+}

E^{0} = 0.153 V

E^{0}

C u^{+} + e^{-} ⟶ C u

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

A g^{+} (a q) + e^{-} \to A g (s); E^{0} = 0.80 V

2 H^{+} (a q) + 2 e^{-} \to H_{2} (g); E^{0} = 0 V

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

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