Ag(s)/Ag+(0.30M)//Ag+(1.5M)/Ag(s) at anode oxidation Ag(s) ------------------- -
ID: 699412 • Letter: A
Question
Ag(s)/Ag+(0.30M)//Ag+(1.5M)/Ag(s)
at anode oxidation Ag(s) ------------------- - Ag+(aq) + e-
at cathode reduction Ag+(aq) + e- ----------------- Ag(s)
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Ag(s) + Ag+(aq, 1.5) ---------- Ag+(0.30M) + Ag(s)
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E0 cell = 0.0V
Ecell= E0cell - 0.059/n log[Ag+] at anode/[Ag+] at cathode
= 0.0- 0.059/1 log[0.30/1.5]
Ecell= -0.059 x ( -0.699)
Ecell= 0.041 V
Ecell = 0.04V
The answer is B.
Explanation / Answer
7. What is the cell potential at 298 K for the electrochemical cell diagrammed below? ) 0.02 V B) +0.04 V C)0.36 V D) +0.02 V E) 0.00 V