In the electrochemical cell shown above, which of the following half-reactions occurs at the anode?

{C u}^{2 +} + e \to {C u}^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $Z n (s) \to {Z n}^{2 +} + 2 e$
At anode, oxidation will take place and $Z n (s)$ will get oxidised to $Z n^{2 +}$ leaving negatively charged electrons, making anode negatively charged, repelling negatively charged $N O_{3}^{-}$ .

Suggest Corrections

Similar questions

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)

The cell representation of the given reaction is:
Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s)

Z n ∣ ∣ {Z n}^{2 +} (a q) ∣ ∣ ∣ ∣ {C u}^{2 +} (a q) ∣ ∣ C u

{C u}^{2 +} + {2 e}^{-} ⟶ C u, E^{\circ} = + 0.35 V

{Z n}^{2 +} + {2 e}^{-} ⟶ Z n, E^{\circ} = - 0.763 V

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

C u I (s) + e^{-} \to C u (s) + I^{-};

E_{C u^{+} / C u}^{o} = - 0.16 V

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s);

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

Join BYJU'S Learning Program