Neet Solved Paper 2013 Question 8
Question: A button cell used in watches functions as following
$ Zn(s)+Ag _2O(s)+H _2O(l)2Ag(s) $ $ +Z{n^{2+}}(aq)+2O{H^{-}}(aq) $ If half-cell potentials are $ Z{n^{2+}}(aq)+2{e^{-}}\to Zn(s)E^{o}=-0.76V $ $ Ag _2O(s)+H _2O(l)+2{e^{-}} $ $ \to 2Ag(s)+2O{H^{-}}(aq), $ $ E^{o}=0.34V $ The cell potential will be
Options:
A) 1.10 V
B) 0, 42 V
C) 0.84 V
D) 1.34 V
Show Answer
Answer:
Correct Answer: A
Solution:
- Anode is always the site of oxidation thus anode half-cell is
$ Z{n^{2+}}(aq)+2{e^{-}}\xrightarrow[{}]{{}}Zn(s);E^{o}=-0.76,V $
Cathode half-cell is $ Ag _2O(s)+H _2O(l)+2{e^{-}}\xrightarrow[{}]{{}} $ $ 2Ag(s)+2O{H^{-}}(ag);E^{o}=0.34,V $ $ E^o _{cell}=E^o _{cathode}-E^o _{anode} $ $ =0.34-(-0.76)=+1.10V $
