A) 1.54 V
B) ? 1.54 V
C) ? 0.19 V
D) + 0.19 V
Correct Answer: C
Solution :
\[F{{e}^{2+}}+Zn\to Z{{n}^{2+}}+Fe\] \[EMF={{E}_{\text{cathode}}}-{{E}_{\text{anode}}}\]\[{{H}^{+}}(0.025\,M)\to {{H}^{+}}({{10}^{-8}}\,M)\] \[EMF=-0.19\,V\].You need to login to perform this action.
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