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
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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
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