A) \[-1.532V\]
B) \[-1.503V\]
C) \[1.532V\]
D) \[-3.06V\]
Correct Answer: A
Solution :
From the given cell, the cell reaction is \[2Ag(s)+Z{{n}^{2+}}(0.1\,M)\xrightarrow{{}}\] \[2A{{g}^{+}}(0.1\,M)+Zn(s)\] The Nernst equation is \[{{E}_{cell}}={{E}^{o}}_{cell}-\frac{0.0591}{n}\log \frac{{{[A{{g}^{+}}]}^{2}}}{[Z{{n}^{2+}}]}\] or, \[{{E}_{cell}}=(-1.562)-\frac{(0.0591)}{2}\log \frac{{{(0.1)}^{2}}}{(0.1)}\] (where, \[{{E}^{o}}_{cell}=-1.562\,V\]) or \[{{E}_{cell}}=(-1.562)-\frac{0.0591}{2}\log {{10}^{-1}}\] \[=-1.562+\frac{0.0591}{2}\] \[=-1.562+0.02955=-1.532\,\text{V}\]
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