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Electrochemical Series

Divide the fo...

Question

Divide the following redox reactions into oxidation and reduction half reactions.
Zn + $P b^{2} +$ → $Z n^{2}$ $^{+}$ + Pb

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Solution

The redox reactions into oxidation and reduction half reactions as follows:
Oxidation half reaction : Zn → $Z n^{2}$ $^{+}$ + 2e- (Involves the loss of two electrons)
Reduction half reaction : $P b^{2}$ $^{+}$ + 2e- → Pb (Involves the gain of two electrons)

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Similar questions

Divide the following redox reactions into oxidation and reduction half reactions.

(i) $Z n + P b^{2} \to Z n^{2} + P b$

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

(iii) $C l_{2} + 2 B r^{-} \to B r_{2} + 2 C l^{-}$

Q. Divide the following redox reactions into oxidation and reduction half reactions.

Z n + C u^{2}^{+} \to Z n^{2}^{+} + C u

Q. The redox reaction

Z n (s) + P b {C l}_{2} (a q) ⟶ P b (s) + Z n {C l}_{2} (a q)

can be represented as following half equations :

Z n (s) ⟶ {Z n}^{2 +} (a q) + 2 e^{-}

(oxidation) and

{P b}^{2 +} (a q) + 2 e^{-} ⟶ P b (s)

(reduction).
If true enter 1, else enter 0.

Q. Name the oxidant and reductant in the following redox reaction. Zn + pbcl2 = Pb +ZnCl2

Q. Calculate EMF of the cell given: Zn(s) / Zn²⁺ (0.1M) // Pb²⁺ (0.02M) / Pb(s). Also write the chemical reaction at the anode & cathode. (E⁰ Zn²⁺/Zn = -0.76V, E⁰ Pb²⁺/Pb = -0.13V)

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Electrochemical Series

Standard XII Chemistry

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