A) ? 1.20 V
B) + 0.32 V
C) ? 0.32 V
D) + 1.20 V
Correct Answer: B
Solution :
\[F{{e}^{+2}}+Zn\to Z{{n}^{2+}}+Fe\] \[EMF={{E}_{\text{cathode}}}-{{E}_{\text{anode}}}\]\[=0.44-(0.76)=+0.32\,V\].You need to login to perform this action.
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