one half cell in a voltaic cell is constructed from a silver wire dipp

1.	one half cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solutions of unknown conc. it's other half cell consists of a wire electrode dipping in 0.1m solution of zn(no3)2.Avoltage of 1.48v is measured for this cell
Answer» *Ag+ = 1.247 10^-2 M Explanation:** We know that E^o (Zn2+ /Zn) = -0.763 V and E^o (Ag2+ /Ag) = +0.80 V In Electrochemical cell, Zn(s) [Zn2+(0.10M] [Ag+ Conc.] Ag(s) E^o (cell) = E^o(R) - E^o(L) E^o (cell) = 0.80 (-0.763) V E^o (cell) = 1.563 V Also E^o (cell) = [E^o(R) - E^o(L)] - 0.0591/n. Log [Zn2+]/ [Ag+]^2 1.48 = 1.563 - 0.0591/2 Log [0.10]/ [Ag+]^2 Log[0.10]/ [Ag+]^2 = 0.083/0.02955 Log[0.10]/ [Ag+]^2 = 2.8087 Therefore [0.10]/ [Ag+]^2 = Antilog of 2.8087 [0.10]/ [Ag+]^2 = 643.7 [Ag+]^2 = 0.10 / 643.7 [Ag+]^2 = 1.553 * 10^-4 Hence Ag+ = ROOT of 1.553 * 10^-4 *Ag+ = 1.247 10^-2 M**

one half cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solutions of unknown conc. it's other half cell consists of a wire electrode dipping in 0.1m solution of zn(no3)2.Avoltage of 1.48v is measured for this cell

Answer»

Explanation:

We know that E^o (Zn2+ /Zn) = -0.763 V

and E^o (Ag2+ /Ag) = +0.80 V

In Electrochemical cell,

Zn(s) [Zn2+(0.10M] [Ag+ Conc.] Ag(s)

E^o (cell) = E^o(R) - E^o(L)

E^o (cell) = 0.80 (-0.763) V

E^o (cell) = 1.563 V

Also E^o (cell) = [E^o(R) - E^o(L)] - 0.0591/n. Log [Zn2+]/ [Ag+]^2

1.48 = 1.563 - 0.0591/2 Log [0.10]/ [Ag+]^2

Log[0.10]/ [Ag+]^2 = 0.083/0.02955

Log[0.10]/ [Ag+]^2 = 2.8087

Therefore [0.10]/ [Ag+]^2 = Antilog of 2.8087

[0.10]/ [Ag+]^2 = 643.7

[Ag+]^2 = 0.10 / 643.7

[Ag+]^2 = 1.553 * 10^-4

Hence Ag+ = ROOT of 1.553 * 10^-4

Ag+ = 1.247 * 10^-2 M