Problem: O2(g)+4H+(aq)+2Zn(s)→2H2O(l)+2Zn2+(aq)Indicate the half-reaction occurring at Cathode.

FREE Expert Solution

lose electrons → oxidation → anode

gain electrons → reduction → cathode


O2(g)+ 4 H+(aq)+2 Zn(s)→2 H2O(l)+2 Zn2+(aq)

 
Analyze the change in charge of each element: 

Zn(s)→ Zn2+(aq)

  • 0 → +2 charge
  • Lost electrons → anode

O2(g) → 2 H2O(l)

  • 0 → -2 charge
  • gained electrons → cathode
  • H is included in H2O so we also include the H+ on the reactant side to balance. 
  • We also add electrons to balance the positive charge. 


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Problem Details

O2(g)+4H+(aq)+2Zn(s)→2H2O(l)+2Zn2+(aq)

Indicate the half-reaction occurring at Cathode.

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