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.
write the balanced equation for the overall reaction in acidic solution.

Explanation:

Step1: Determine the oxidation and reduction half - reactions

The \(Cu^{2+}/Cu\) half - reaction has a higher \(E^{\circ}_{red}\) value (\(0.337\ V\)) compared to the \(AgCl/Ag\) half - reaction (\(0.2223\ V\)). So, \(Cu^{2+}\) will be reduced and \(Ag\) (from \(AgCl\)) will be oxidized.
The reduction half - reaction is \(Cu^{2+}(aq)+2e^{-}\to Cu(s)\)
The oxidation half - reaction is \(2Ag(s)+2Cl^{-}(aq)\to 2AgCl(s)+2e^{-}\) (multiplied by 2 to balance the electrons)

Step2: Combine the half - reactions

Add the reduction and oxidation half - reactions together:
\(Cu^{2+}(aq)+2Ag(s)+2Cl^{-}(aq)\to Cu(s)+2AgCl(s)\)

Answer:

\(Cu^{2+}(aq)+2Ag(s)+2Cl^{-}(aq)\to Cu(s)+2AgCl(s)\)