A) + 0.04V
B) -0.04V
C) +1.54 V
D) -1.54V
Correct Answer: C
Solution :
Key Idea: (i) Use the formula \[E_{cell}^{o}=E_{C}^{o}-E_{A}^{o}\]. (ii) Find the cathode and anode by cell reaction and then substitute the values in terms of reduction potential. \[Zn/Z{{n}^{2+}}||\,A{{g}^{+}}/Ag\] \[\therefore \] Zn is anode and Ag is cathode. Given, \[{{E}^{o}}_{Z{{n}^{2+}}/Zn}=-0.76\,V\], \[{{E}^{o}}_{A{{g}^{+}}/Ag}=0.80\,V\] \[\therefore \] \[E_{cell}^{o}=E_{C}^{o}=E_{A}^{o}\] = 0.80 - (- 0.76) = 0.80 + 0.76 = 1.54 VYou need to login to perform this action.
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