| \[Zn(s)+A{{g}_{2}}O(s)+{{H}_{2}}O(l)2Ag(s)\] \[+Z{{n}^{2+}}(aq)+2O{{H}^{-}}(aq)\] |
| \[Z{{n}^{2+}}(aq)+2{{e}^{-}}\to Zn(s){{E}^{o}}=-0.76V\] |
| \[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\] \[\to 2Ag(s)+2O{{H}^{-}}(aq),\]\[{{E}^{o}}=0.34V\] |
A) 1.10 V
B) 0, 42 V
C) 0.84 V
D) 1.34 V
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 \[A{{g}_{2}}O(s)+{{H}_{2}}O(l)+2{{e}^{-}}\xrightarrow[{}]{{}}\] \[2Ag(s)+2O{{H}^{-}}(ag);{{E}^{o}}=0.34\,V\] \[{{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\] \[=0.34-(-0.76)=+1.10\,\,V\]
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