Question

Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) $2C{{r}_{\left( s \right)}}\text{ }+\text{ }3C{{d}^{2+}}_{\left( aq \right)}\to 2C{{r}^{3+}}_{\left( aq \right)}\text{ }+\text{ }3C{{d}_{\left( s \right)}}$
(ii) $F{{e}^{2+}}_{\left( aq \right)}\text{ }+\text{ }A{{g}^{+}}_{\left( aq \right)}\to F{{e}^{3+}}_{\left( aq \right)}\text{ }+\text{ }A{{g}_{\left( s \right)}}$
Calculate the ${{\Delta }_{r}}{{G}^{o}}$, and equilibrium constant of the reactions.

Answer 1

standard state potential of a cell is the potential under standard state conditions, which is approximated with concentrations of 1 molar (1 mole per liter) and pressures of 1 atmosphere at ${{25}^{\circ }}C$(298K). Attempt this question by calculating the standard cell potential and then use them in the formulas written below to calculate ${{\Delta }_{r}}{{G}^{o}}$ and equilibrium constant of the reactions.

Formula used: We will require the following formulas:-
${{\Delta }_{r}}{{G}^{o}}=-nF{{E}^{o}}_{cell} \\\ {{\Delta }_{r}}{{G}^{o}}=-RT\ln K \\\ {{E}^{o}}_{cell}={{E}^{o}}_{R}-{{E}^{o}}_{L} \\\$

Complete step-by-step answer: Let us first begin with the representation of the cell followed by other calculations:-
(i) ${{E}^{o}}_{C{{r}^{3+}}/Cr}=-0.74V$
${{E}^{o}}_{C{{d}^{2+}}/Cd}=-0.40V$
-The galvanic cell of the given reaction is depicted as follows:-
$C{{r}_{(s)}}|C{{r}^{3+}}_{(aq)}||C{{d}^{2+}}_{(aq)}|C{{r}_{(s)}}$
The standard cell potential is ${{E}^{o}}_{cell}={{E}^{o}}_{R}-{{E}^{o}}_{L}$
= -0.40 - (-0.74)
= + 0.34 V
-Calculation of${{\Delta }_{r}}{{G}^{o}}$:-
${{\Delta }_{r}}{{G}^{o}}=-nF{{E}^{o}}_{cell}$
In the equation $2C{{r}_{\left( s \right)}}\text{ }+\text{ }3C{{d}^{2+}}_{\left( aq \right)}\to 2C{{r}^{3+}}_{\left( aq \right)}\text{ }+\text{ }3C{{d}_{\left( s \right)}}$
N (No. of moles of electron) = 6
F (Faraday constant) = 96487 C/mol
${{E}^{o}}_{cell}$ = + 0.34 V
Therefore, ${{\Delta }_{r}}{{G}^{o}}$for the reaction:-
\begin{array}{*{35}{l}} \begin{aligned} & {{\Delta }_{r}}{{G}^{o}}=-nF{{E}^{o}}_{cell}\text{ } \\\ & \text{=}-\text{ }6\text{ }\times \text{ }96487\text{ }C\text{ }mo{{l}^{~-\text{ }1}}~\times \text{ }0.34\text{ }V \\\ \end{aligned} \\\ =\text{ }-\text{ }196833.48\text{ }CV\text{ }mol{{~}^{-\text{ }1}} \\\ =\text{ }-\text{ }196833.48\text{ }J\text{ }mo{{l}^{~-\text{ }1}} \\\ =\text{ }-\text{ }196.83\text{ }kJ\text{ }mo{{l}^{~-\text{ }1}} \\\ \end{array}
-Calculation of Equilibrium constant (K):-
${{\Delta }_{r}}{{G}^{o}}=-RT\ln K$
Where,
R = universal gas constant = $8.314\dfrac{J}{mol\cdot K}$
T= temperature (on Kelvin scale) =298K
Also ${{\Delta }_{r}}{{G}^{o}}=-RT\ln K=-2.303RT\log K$
On rearranging the formula and substituting the values, we get:-
$\begin{aligned} & {{\Delta }_{r}}{{G}^{o}}=-2.303RT\log K \\\ & \log K=\dfrac{-196.83\times {{10}^{3}}}{2.303\times 8.314\times 298} \\\ \end{aligned}$
$\log K=\text{ }34.496$
K = antilog (34.496)
K = $3.13\times {{10}^{-34}}$
-Therefore, the standard cell potential of the given reaction is 0.34V, ${{\Delta }_{r}}{{G}^{o}}=-\text{ }196.83\text{ }kJ\text{ }mo{{l}^{~-\text{ }1}}$and equilibrium constant (K) = $3.13\times {{10}^{-34}}$
(ii) ${{E}^{o}}_{F{{e}^{3+}}/F{{e}^{2+}}}=0.77V$
${{E}^{o}}_{A{{g}^{+}}/Ag}=0.80V$
-The galvanic cell of the given reaction is depicted as follows:-
$F{{e}^{2+}}_{(aq)}|F{{e}^{3+}}_{(aq)}||A{{g}^{+}}_{(aq)}|A{{g}_{(s)}}$
The standard cell potential is ${{E}^{o}}_{cell}={{E}^{o}}_{R}-{{E}^{o}}_{L}$
= 0.80 - 0.77
= 0.03 V
-Calculation of ${{\Delta }_{r}}{{G}^{o}}$:-
${{\Delta }_{r}}{{G}^{o}}=-nF{{E}^{o}}_{cell}$
In the equation $F{{e}^{2+}}_{\left( aq \right)}\text{ }+\text{ }A{{g}^{+}}_{\left( aq \right)}\to F{{e}^{3+}}_{\left( aq \right)}\text{ }+\text{ }A{{g}_{\left( s \right)}}$
N (No. of moles of electron) = 1
F (Faraday constant) = 96487 C/mol
${{E}^{o}}_{cell}$ = + 0.03 V
Therefore, ${{\Delta }_{r}}{{G}^{o}}$for the reaction:-
\begin{array}{*{35}{l}} \begin{aligned} & {{\Delta }_{r}}{{G}^{o}}=-nF{{E}^{o}}_{cell}\text{ } \\\ & \text{=}-\text{ }1\text{ }\times \text{ }96487\text{ }C\text{ }mo{{l}^{~-\text{ }1}}~\times \text{ }0.03\text{ }V \\\ \end{aligned} \\\ =\text{ }-\text{ }2894.61\text{ }CV\text{ }mol{{~}^{-\text{ }1}} \\\ =\text{ }-\text{ }2894.61\text{ }J\text{ }mo{{l}^{~-\text{ }1}} \\\ =\text{ }-\text{ }2.89\text{ }kJ\text{ }mo{{l}^{~-\text{ }1}} \\\ \end{array}
-Calculation of Equilibrium constant (K):-
${{\Delta }_{r}}{{G}^{o}}=-RT\ln K$
Where,
R = universal gas constant = $8.314\dfrac{J}{mol\cdot K}$
T= temperature (on Kelvin scale) =298K
Also ${{\Delta }_{r}}{{G}^{o}}=-RT\ln K=-2.303RT\log K$
On rearranging the formula and substituting the values, we get:-
$\begin{aligned} & {{\Delta }_{r}}{{G}^{o}}=-2.303RT\log K \\\ & \log K=\dfrac{-2.89461\times {{10}^{3}}}{2.303\times 8.314\times 298} \\\ \end{aligned}$
$\log K=\text{ }0.5073$
K = antilog (0.5073)
K = 3.22
-Therefore, the standard cell potential of the given reaction is 0.03V, ${{\Delta }_{r}}{{G}^{o}}=-\text{ }2.89\text{ }kJ\text{ }mo{{l}^{~-\text{ }1}}$and equilibrium constant (K) = 3.22

Note: Kindly remember to accordingly convert units and use them in the formula as required to get the accurate result.
-For such types of long answer questions always prefer the step by step method, so as to avoid any kind of miscalculations and also do not forget negative signs which most of us usually neglect while solving the questions.