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2. Use a table of Standard Electron Reduction Potentials (click the Reference Materials link in the header of this webpage an
3. Identify the half-cell at the cathode (enter a number 1-4) 1. Zn(q) + 2e → Zn* 2. Znis) → Znaq + 2 e 3. Alaq) + 3e > Ale 4
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2. Ans - @ Ans. I. We have Fall = E cathode - E anode Again, E 224 / 2n = -0.76 U and E ALS/AC = -1.66V. So, zn2t is reducedOnidation half reaction :- Al(s) - Al? alt bé . Reduction half reaction :- 2m 2 talt de 20 (1) Balancing the charge and addinBians - Half all are fel felt and Pb/P624 from stans and elutode potential table E P62/Pb = -0.13 V and E Festlife = -0.440.rom an Reduction half: Pb2+ (aq) + 20 → Pb/s). Balancing the charge and atom and adding the equation I fels) + Pb2+ (aq) Fe2+

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