Question

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

• A. $+1.10V$
• B. $-1.10V$
• C. $+0.42V$
• 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, $Z{n}^{2+}/Zn=-0.76V$ $C{u}^{2+}/Cu=+0.34V$ $E_{cell}=E_c-E_a$ $=0.34-(-0.76)$ $=0.34+0.76$ $=1.10V$