Question

The emf of the cell, $ ({{E}_{Z{{n}^{2+}}/Zn}}=-0.76\,V) $ $ Zn/Z{{n}^{2+}}(1M)||C{{u}^{2+}}(1M)Cu $ $ ({{E}_{C{{u}^{2+}}/Cu}}=+0.34V) $ will be

• A. $ +1.10\text{ }V $
• B. $ -1.10\text{ }V $
• C. $ +\text{ }0.42V $
• D. $ -0.42V $

Answer 1

Answer: (A)

Key Idea: (i) Find cathode and anode from given cell reaction. (ii) Calculate emf by formula $ {{E}_{cell}}={{E}_{c}}-{{E}_{a}} $ Given that $ Zn/Z{{n}^{2+}}||C{{u}^{2+}}/Cu $ $ \therefore $ Zn is anode and Cu is cathode Given, $ Z{{n}^{2+}}/Zn=-0.76\,V $ $ C{{u}^{2+}}/Cu=+0.34\,V $ $ {{E}_{cell}}={{E}_{c}}-{{E}_{a}} $ $ =0.34-(-0.76) $ $ =0.34+0.76 $ $ =1.10V $