[Ag^+] Â is 3.617M .
Given ,
Calomel : Hg2Cl2 (s) +2e^- ==> 2Hg(l) +2Cl^- (aq) Â , E^0 = 0.256V
Silver: Â Ag+ (aq) +e^- ==>Ag(s) Â , E^- = 0.806V
We know E^0 cell = E^0 cathode - E^0 anode
=> E^0cell = E^0 Ag/Ag^+ Â - Â E^0 Hg2Cl2/Cl^-
=> E^0cell = 0.806 - 0.256 = 0.55Volts
Now ,according to Nernst equation ,
Ecell = E^0cell - 0.0592/n log [product / reactant  ]
Ecell = E^0cell - 0.0592/2 log { [Ag^+]2/[Cl^-]2}
0.517= 0.55 - 0.0592/2 log {[Ag^+]2/ 1}
0.033 = 0.0592/2 (2 log [Ag^+] )
log[Ag^+] = 0.033/0.0592
taking antilog on both sides ,
[Ag^+] = 3.617 M
Hence , [Ag^+] is 3.617M .
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Disclaimer :incomplete and incorrect question .here is the complete question .
Question : A chemist designs an ion-specific probe for measuring [Ag^+] in an NaCl solution saturated with AgCl.one half-cell has an Ag wire electrode immersed in the unknown AgCl-saturated NaCl solution . it is connected through a salt bridge to the other half-cell ,which has a calomel reference electrode [ a platinum wire immersed in a paste of mercury and calomel (Hg2Cl2) ] in a saturated KCl solution .
Amining engineer wants an ore sample analyzed with Ag+ selective probe . after penetrating the ore sample , the chemist measures the cell voltage as 0.517V . what is [Ag^+] ?
Calomel : Hg2Cl2 (s) +2e^- ==> 2Hg(l) +2Cl^- (aq) Â , E^0 = 0.256V
Silver: Â Ag+ (aq) +e^- ==>Ag(s) Â , E^- = 0.806V
(Hint: assume that [Cl^-] is so high that it is essentially constant, use 1.00 M )
record your answer in molarity with 3 decimals .
