A) >80 ml
B) < 80 ml
C) = 80 ml
D) \[\text{2}.\text{3 }\times \text{1}{{0}^{-\text{16}}}\text{ mole}/\text{litre}\]
Correct Answer: B
Solution :
As A has more £°red value than B, A will act as cathode in the galvanic cell. Hence, \[E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\] \[=\left( \text{2}.\text{23} \right)-\left( -\text{1}.\text{43} \right)\] \[=\text{2}.\text{23 }+\text{1}.\text{43 }=\text{3}.\text{66 V}\]
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