Question

a student wishes to determine the chloride ion concentration in a water sample at 25 °c using a galvanic cell constructed with a graphite electrode and a half-cell of
agcl(s) + e⁻ → ag(s) + cl⁻(aq) e°red = 0.2223 v
and a copper electrode with 0.500 m cu²⁺ as the second half cell
cu²⁺(aq) + 2 e⁻ → cu(s) e°red= 0.337 v
the measured cell potential when the water sample was placed into the silver side of the cell was 0.0925 v.
what is the standard cell potential for this cell in v?

Explanation:

Step1: Determine the anode and cathode

The half - reaction with the lower standard reduction potential will be the anode (oxidation occurs there). Since \(E^{\circ}_{red}\) for \(AgCl/Ag\) (\(0.2223\ V\)) is less than \(E^{\circ}_{red}\) for \(Cu^{2+}/Cu\) (\(0.337\ V\)), the \(AgCl/Ag\) half - cell is the anode (oxidation: \(Ag(s)+Cl^{-}(aq)\to AgCl(s)+e^{-}\), \(E^{\circ}_{ox}=- 0.2223\ V\)) and the \(Cu^{2+}/Cu\) half - cell is the cathode (reduction: \(Cu^{2+}(aq)+2e^{-}\to Cu(s)\), \(E^{\circ}_{red}=0.337\ V\))

Step2: Calculate the standard cell potential

The formula for the standard cell potential \(E^{\circ}_{cell}\) is \(E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}\)
Multiply the \(AgCl/Ag\) half - reaction by \(2\) to balance the electrons (\(2Ag(s)+2Cl^{-}(aq)\to 2AgCl(s)+2e^{-}\), \(E^{\circ}_{ox}=- 0.2223\ V\)) and keep the \(Cu^{2+}/Cu\) half - reaction as \(Cu^{2+}(aq)+2e^{-}\to Cu(s)\), \(E^{\circ}_{red}=0.337\ V\)
\(E^{\circ}_{cell}=E^{\circ}_{red}(Cu^{2+}/Cu)-E^{\circ}_{red}(AgCl/Ag)\)
\(E^{\circ}_{cell}=0.337-0.2223\)

Answer:

\(E^{\circ}_{cell}=0.1147\ V\)