Question

question 19.b of 20
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.
given the standard cell potential for this cell is 0.115 v, what is the value of the standard free energy (in kj) for this reaction?

Explanation:

Step1: Identify the formula

The formula for standard free energy change is $\Delta G^{\circ}=-nFE^{\circ}_{cell}$, where $n$ is the number of moles of electrons transferred, $F = 96485\ C/mol$ (Faraday's constant), and $E^{\circ}_{cell}$ is the standard cell potential.

Step2: Determine the value of $n$

For the reaction:
The half - reactions are:
$2AgCl(s)+2e^{-}\to2Ag(s)+2Cl^{-}(aq)$ (reduction)
$Cu(s)\to Cu^{2 +}(aq)+2e^{-}$ (oxidation)
The overall reaction is $2AgCl(s)+Cu(s)\to2Ag(s)+2Cl^{-}(aq)+Cu^{2 +}(aq)$. The number of moles of electrons transferred $n = 2$.

Step3: Substitute the values into the formula

Given $E^{\circ}_{cell}=0.115\ V$, $n = 2$, and $F=96485\ C/mol$
$\Delta G^{\circ}=-nFE^{\circ}_{cell}$
$\Delta G^{\circ}=-(2\ mol)(96485\ C/mol)(0.115\ V)$
First, calculate $(2)(96485)(0.115)=2\times96485\times0.115 = 2\times11095.775=22191.55\ J$
Since $1\ kJ = 1000\ J$, $\Delta G^{\circ}=- 22.2\ kJ$

Answer:

$-22.2\ kJ$