| A button cell used in watches functions as following [NEET 2013] |
| \[Zn(s)+A{{g}_{2}}O(s)+{{H}_{2}}O(l)2Ag(s)\]\[+Z{{n}^{2+}}(aq)+2O{{H}^{-}}(aq)\] |
| If half cell potentials are |
| \[Z{{n}^{2+}}(aq)+2{{e}^{-}}\to Zn(s){{E}^{o}}=-0.76V\]\[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\]\[\to 2Ag(s)+2O{{H}^{-}}(aq),\]\[{{E}^{o}}=0.34V\] |
| The cell potential will be |
A) 1.10 V
B) 0, 42 V
C) 0.84 V
D) 1.34 V
Correct Answer: A
Solution :
| [a] 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 |
| \[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\xrightarrow[{}]{{}}\] |
| \[2Ag(s)+2O{{H}^{-}}(ag);{{E}^{o}}=0.34\,V\] |
| \[{{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\] |
| \[=0.34-(-0.76)=+1.10\,\,V\] |
You need to login to perform this action.
You will be redirected in
3 sec
