Question

13. A zinc electrode is submerged in an acidic 0.40 M Zn? solution which is connected by a salt bridge to a 1.50 M Ag solution containing a silver electrode. Determine the initial voltage of the cell at 298K 32

Answer 1

Q13. Reduction half reaction at cathode : Ag⁺ (aq) + e^-$\rightarrow$ Ag (s) : E^o_red = 0.80 V

Oxidation half reaction at anode : Zn (s) $\rightarrow$ Zn²⁺ (aq) + 2 e^- : E^o_ox = 0.76 V

Overall reaction : Zn (s) + 2 Ag⁺ (aq) $\rightarrow$ Zn²⁺ (aq) + 2 Ag (s)

Standard cell potential, E^o_cell = E^o_red + E^o_ox

E^o_cell = 0.80 V + 0.76 V

E^o_cell = 1.56 V

According to Nernst equation,

E_cell = E^o_cell - (0.0592 V / n) * log([Zn²⁺] / [Ag⁺]²)

where n = number of electrons transferred in overall equation = 2

E_cell = 1.56 V - (0.0592 V / 2) * log(0.40 M / (1.50 M)²)

E_cell = 1.56 V - (0.0592 V / 2) * (-0.75)

E_cell = 1.56 V + 0.022

E_cell = 1.582 V

Initial voltage of cell = 1.582 V