Question

Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) $2Cr(s)+3C{d}^{2+}(aq)\to 2C{r}^{3+}(aq)+3Cd$
(ii) $F{e}^{2+}(aq)+A{g}^{+}(aq)\to F{e}^{3+}(aq)+Ag(s)$
Calculate the ${\mathrm{\Delta }}_r{G}^{\mathrm{\Theta }}$ and equilibrium constant of the reactions

Answer 1

Hint: The standard cell potential, ${\mathrm{\Delta }}_r{G}^{\mathrm{\Theta }}$and equilibrium constant of the reactions can be calculated using standard electrode potentials of the elements involved. The electrode potential cannot be obtained empirically. It is calculated using a reference hydrogen electrode.

Complete step by step solution:
Standard electrode potential $(E{}^{\circ })$ is defined as the measures the individual potential of reversible electrode at standard state with ions at an effective concentration of 1mol $d{m}^{-3}$at the pressure of 1 atm.
(i) $2Cr(s)+3C{d}^{2+}(aq)\to 2C{r}^{3+}(aq)+3Cd$
The formula of calculating cell potential is :
$${E^{{}^{\circ }}}_{cell}={E^{{}^{\circ }}}_{cathode}-{E^{{}^{\circ }}}_{anode}$$
In the equation, cathode is cadmium with electrode potential , -0.40V, anode is chromium with electrode potential, -0.74V. Therefore, substituting the value in the above formula,
$$\begin{array}{rl}& {E^{{}^{\circ }}}_{cell}=-0.40-(-0.74)\\ & {E^{{}^{\circ }}}_{cell}=0.34V\end{array}$$
In a galvanic cell, where a spontaneous redox reaction drives the cell to produce an electric potential, Gibbs free energy ${\mathrm{\Delta }}_r{G}^{\mathrm{\Theta }}$ must be negative, in accordance with the following equation:
$$\mathrm{\Delta }G{{}^{\circ }}_{cell}=-nFE{{}^{\circ }}_{cell}$$
Since we know the electrode potential, and number electrons involved in the reaction is 6 and value of faraday constant is 96500, Therefore, substituting the value in the above formula,
$$\begin{array}{rl}& \mathrm{\Delta }G{{}^{\circ }}_{cell}=-6\times 96500\times 0.34\\ & \mathrm{\Delta }G{{}^{\circ }}_{cell}=-196860J/mol=-196.86kJ/mol\end{array}$$
The formula for equilibrium constant is,
$$\log K_c={\displaystyle \frac{n{E_{cell}}^o}{0.059}}={\displaystyle \frac{6\times 0.34}{0.059}}=34.576$$
$$K_c=3.76\times {10}^{34}$$
Equilibrium constant is $$3.76\times {10}^{34}$$.
(ii) $F{e}^{2+}(aq)+A{g}^{+}(aq)\to F{e}^{3+}(aq)+Ag(s)$
In the equation, cathode is silver with electrode potential , 0.80V, anode is iron with electrode potential, 0.77V. Therefore, substituting the value in the above formula,
$$\begin{array}{rl}& {E^{{}^{\circ }}}_{cell}=0.80-(0.77)\\ & {E^{{}^{\circ }}}_{cell}=0.03V\end{array}$$
Now Gibbs free energy will be:
$$\begin{array}{rl}& \mathrm{\Delta }G{{}^{\circ }}_{cell}=-1\times 96500\times 0.03\\ & \mathrm{\Delta }G{{}^{\circ }}_{cell}=-2895J/mol=-2.895kJ/mol\end{array}$$
The equilibrium constant is,
$$\log K_c={\displaystyle \frac{n{E_{cell}}^o}{0.059}}={\displaystyle \frac{1\times 0.03}{0.059}}=0.508$$
$$K_c=3.22$$

Note: The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.