A) increase of 59 mV
B) decrease of 59 mV
C) increase of 29.5 mV
D) decrease of 29.5 mV
Correct Answer: B
Solution :
Key Idea: \[E=\frac{0.059}{n}\log \frac{1}{C}\] where n = no. of electrons in electrode in reaction \[C=\] concentration \[n=2\,(for\,Z{{n}^{2+}}\to Zn+2{{e}^{-}})\] \[C=1\] for (initial concentration) when \[C=100\] \[{{E}_{Z{{n}^{2+}}/Zn}}=\frac{0.059}{2}\log \frac{1}{100}\] or \[{{E}_{Z{{n}^{2+}}/Zn}}=\frac{0.059}{2}\times -2\] \[\therefore \] \[{{E}_{Z{{n}^{2+}}/Zn}}=0.059\] \[\therefore \] If \[Z{{n}^{2+}}/Zn\] electrode is diluted to 100 times the emf decrease by 59 mV.
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