A) + 0.34V
B) + 0.76V
C) -1.1 V
D) +1.1 V
Correct Answer: D
Solution :
As the \[{{\text{E}}^{\text{o}}}\text{re}{{\text{d}}^{\text{0}}}\text{.}\]is lesser for Zn, hence it will act as anode and Cu as cathode in the galvanic cell. Hence, \[\text{E}_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] = (0.34) - (- 0.76) =1.1V
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