A) 0.42 V
B) - 0.42 V
C) - 1.1 V
D) 1.1 V
Correct Answer: A
Solution :
Key Idea : (i) Decide cathode and anode (ii) \[{{E}^{o}}_{cell}={{E}^{o}}_{C}-{{E}^{o}}_{A}\] when values of electrode potential are substituted in terms of reduction potential. Given, \[E_{Zn/Z{{n}^{2+}}}^{o}=0.76\,\,V\] \[E_{Cu/C{{u}^{2+}}}^{o}=0.34\,\,V\] \[\therefore \] Zn is anode (\[\because \] it has higher oxidation potential) \[\therefore \] \[E_{Z{{n}^{2+}}/Zn}^{o}=-0.76\,\,V\] and \[E_{C{{u}^{2+}}/Cu}^{o}=-0.34\,\,V\] \[{{E}^{o}}_{cell}={{E}^{o}}_{cell}-{{E}^{o}}_{A}\]You need to login to perform this action.
You will be redirected in
3 sec