Question

At 25°C, the emf of the following electrochemical cell.

\[\ce{Ag_{(s)} | Ag^+ (0.01 M) | | Zn^{2+} {(0.1 M)} | Zn_{(s)}}\] will be:

(Given \[\ce{E^0_{cell}}\] = −1.562 V)

Options

• −1.432 V

• −1.4732 V

• +1.432 V

• −1.4436 V

Answer 1

−1.4732 V

Explanation:

For the given electrochemical cell, the oxidation reaction at anode, reduction reaction at cathode and overall cell reaction are as follows:

\[\ce{2Ag -> 2Ag^+ + 2e^-}\] (Oxidation at anode)

\[\ce{Zn^{+2} + 2e^- -> Zn}\] (Reduction at cathode)

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\[\ce{2Ag + Zn^{+2} -> 2Ag^+ + Zn}\] (Overall cell reaction)

∴The emf of the cell is given by

E_cell = `"E"_"cell"^0 - 0.0592/"n" log_10  (["Ag"^+]^2)/(["Zn"^(+2)])`

Substituting the values in above equation,

E_cell = `-1.562 - 0.0592/2 log_10  (0.01)^2/((0.1))`

= `-1.562 - 0.0592/2 log_10 10^-3`

= `-1.562 - (-3) 0.0592/2`

= −1.562 + 0.0888

= −1.4732 V