| [A] Emf of cell = (oxidation potential of anode) (reduction potential of cathode) |
| [B] Emf of cell = (oxidation potential of anode) + (reduction potential of cathode) |
| [C] Emf of cell = (reduction potential of anode) + (reduction potential of cathode) |
| [D] Emf of cell = (oxidation potential of anode)-(0xidation potential of cathode) |
A) [C] and [A]
B) [A] and [B]
C) [C] and [D]
D) [B] and [D]
Correct Answer: D
Solution :
\[{{E}_{cell}}={{E}^{o}}_{_{(red)}^{cathode}}-{{E}^{o}}_{_{(red)}^{anode}}\] or \[{{E}_{cell}}={{E}^{o}}_{_{(red)}^{cathode}}+{{E}^{o}}_{_{(oxid)}^{Anode}}\] or \[{{E}_{cell}}={{E}^{o}}_{_{(oxid)}^{anode}}-{{E}^{o}}_{_{(oxid)}^{cathode}}\]
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