A) 0,177V
B) - 0.3V
C) 0.052V
D) 0.104V
Correct Answer: A
Solution :
Key Idea: For concentration cells \[E_{cell}^{o}=\frac{0.0591}{1}\log \left[ \frac{anode}{cathode} \right]\] The electrode having lower concentration acts as anode and the electrode having higher concentration acts as cathode. The cells having same electrolytes at cathode and anode are called concentration cells. Given: \[pH=3\] for first solution and \[pH=6\] for second solution. \[[{{H}^{+}}]\] for first solution \[={{10}^{-pH}}={{10}^{-3}}\] \[[{{H}^{+}}]\] for second solution \[={{10}^{-pH}}={{10}^{-6}}\] \[\therefore \] \[E_{cell}^{o}=-\frac{0.0591}{1}\log \left[ \frac{{{10}^{-6}}}{{{10}^{-3}}} \right]\] \[=0.091\,\,\,\log \,{{10}^{-3}}\] \[=-0.0591\times (-3\log 10)\] \[=-0.0591\times (-3\times 1)\] \[=-0.0591-3\] \[\therefore \] \[E_{cell}^{o}=0.177\,\,V\]
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