One half cell in a voltaic cell is constructed by dipping silver rod in AgNO3 solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO4. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag+ ions used in terms of log x (x = [Ag+])? E Zn2+/Zn = −0.76 V, E Ag+/Ag = +0.80 V, 2.303RTF = 0.059 V
- 23.9
- 45.9
- 2.92
- 5.94
Answer
(2) 45.9
Solution
Zn has the lower reduction potential ⇒ Zn is the anode, Ag is the cathode.
Cell reaction: Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s), n = 2
E cell = 0.80 − (−0.76) = 1.56 V
Nernst equation: Ecell = E cell - 0.0592 [ Zn2+][Ag+]2
⇒ 1.60 = 1.56 - 0.0592 1x2
∵ 1x2 = -2 x
⇒ 1.60 = 1.56 + 0.059 log x
⇒ log x = 0.040.059 = 45.9
Answer: (2)