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 ${{\Delta }_{r}}{{G}^{\Theta }}$ and equilibrium constant of the reactions

Answer 1

The standard cell potential, ${{\Delta }_{r}}{{G}^{\Theta }}$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,

& {{E}^{{}^\circ }}_{cell}=-0.40-(-0.74) \\\ & {{E}^{{}^\circ }}_{cell}=0.34V \\\ \end{aligned}$$ In a galvanic cell, where a spontaneous redox reaction drives the cell to produce an electric potential, Gibbs free energy ${{\Delta }_{r}}{{G}^{\Theta }}$ must be negative, in accordance with the following equation: $$\Delta G{{{}^\circ }_{cell}}~=\text{ }-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{aligned} & \Delta G{{{}^\circ }_{cell}}~=\text{ }-6\times 96500\times 0.34 \\\ & \Delta G{{{}^\circ }_{cell}}~=-196860J/mol=-196.86kJ/mol \\\ \end{aligned}$$ The formula for equilibrium constant is, $$\log {{K}_{c}}=\dfrac{n{{E}_{cell}}^{o}}{0.059}=\dfrac{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{aligned} & {{E}^{{}^\circ }}_{cell}=0.80-(0.77) \\\ & {{E}^{{}^\circ }}_{cell}=0.03V \\\ \end{aligned}$$ Now Gibbs free energy will be: $$\begin{aligned} & \Delta G{{{}^\circ }_{cell}}~=\text{ }-1\times 96500\times 0.03 \\\ & \Delta G{{{}^\circ }_{cell}}~=-2895J/mol=-2.895kJ/mol \\\ \end{aligned}$$ The equilibrium constant is, $$\log {{K}_{c}}=\dfrac{n{{E}_{cell}}^{o}}{0.059}=\dfrac{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.