Solveeit Logo
Login

Question

Question: Calculate the equilibrium constant of the reaction at \( 298K. \) \( M{g_{\left( s \right)}}{\te...

Calculate the equilibrium constant of the reaction at 298K.298K.
Mg(s) + 2Ag(aq)+→ Mg(aq) 2++2Ag(s); Ecello=+3.16VM{g_{\left( s \right)}}{\text{ }} + {\text{ }}2Ag_{\left( {aq} \right)}^ + \to {\text{ }}Mg_{\left( {aq} \right){\text{ }}}^{^{2 + }} + 2A{g_{\left( s \right)}};{\text{ }}E_{cell}^{\text{o}} = + 3.16V

Explanation

Solution

Hint : The equilibrium constant, denoted as "K", represents the relationship between the reactant and product concentration of a reaction. To calculate the equilibrium constant of the given reaction taking place in a galvanic cell involving magnesium electrodes and silver electrodes (as given in question), we need to use the given value of Ecello,E_{cell}^{\text{o}}, by which we can calculate the Standard Gibbs energy change (ΔGo).\left( {\Delta {G^{\text{o}}}} \right). Once that is calculated, we need to substitute value for ΔGo\Delta {G^{\text{o}}} in the equation relating K and ΔGo,\Delta {G^{\text{o}}}, which will give the required product.

Complete Step By Step Answer:
The Gibbs energy change accompanying a cell reaction is related to the EMF of the cell reaction by:
ΔG=−nFEcell\Delta G = - nF{E_{cell}}
Where 'n' is the number of moles of electrons transferred in the cell reaction, 'F' is the quantity of electricity passed through the cell for a cell reaction involving the transfer of one mole of electrons (1F=1Faraday=96500C).\left( {1F = 1Faraday = 96500C} \right).
Since we are given the value of standard EMF Ecello,E_{cell}^{\text{o}}, the standard Gibbs energy of the reaction (ΔGo)\left( {\Delta {G^{\text{o}}}} \right) is given as:
ΔGo=−nFEcello→(1)\Delta {G^{\text{o}}} = - nFE_{cell}^{\text{o}} \to (1)
The thermodynamic relation that connects the standard Gibbs energy change (ΔGo)\left( {\Delta {G^{\text{o}}}} \right) to the equilibrium constant K is:
ΔGo=−2.303RTlog⁡K\Delta {G^{\text{o}}} = - 2.303RT\log K
⇒log⁡K=−ΔGo2.303RT→(2)\Rightarrow \log K = - \dfrac{{\Delta {G^{\text{o}}}}}{{2.303RT}} \to (2)
⇒K=antilog(−ΔGo2.303RT)\Rightarrow K = {\text{antilog}}\left( { - \dfrac{{\Delta {G^{\text{o}}}}}{{2.303RT}}} \right)
We are given that:
Mg(s) + 2Ag(aq)+→ Mg(aq) 2++2Ag(s) M{g_{\left( s \right)}}{\text{ }} + {\text{ }}2Ag_{\left( {aq} \right)}^ + \to {\text{ }}Mg_{\left( {aq} \right){\text{ }}}^{^{2 + }} + 2A{g_{\left( s \right)}}{\text{ }}
Where Ecello=+3.16V;T=298K;n=2(from reaction)E_{cell}^{\text{o}} = + 3.16V;T = 298K;n = 2({\text{from reaction}})
Substituting this value in (1),(1), we get:
ΔGo=−nFEcello\Delta {G^{\text{o}}} = - nFE_{cell}^{\text{o}}
=−2×96500Cmol−1×3.16V= - 2 \times 96500Cmo{l^{ - 1}} \times 3.16V
=−609880 Jmol−1= - 609880\,Jmo{l^{ - 1}}
Substituting the value of ΔGo\Delta {G^{\text{o}}} in (2),(2), we get:
log⁡K=−ΔGo2.303RT\log K = - \dfrac{{\Delta {G^{\text{o}}}}}{{2.303RT}}
log⁡K=−−609880Jmol−12.303×8.314JK−1mol−1×298K\log K = - \dfrac{{ - 609880Jmo{l^{ - 1}}}}{{2.303 \times 8.314J{K^{ - 1}}mo{l^{ - 1}} \times 298K}}
log⁡K=106.8868\log K = 106.8868
⇒K=antilog(106.8868)\Rightarrow K = {\text{antilog}}(106.8868)
K=7.7054×10106K = 7.7054 \times {10^{106}}
Hence, the equilibrium constant for the given reaction is K=7.7054×10106.K = 7.7054 \times {10^{106}}.

Note :
An alternate method to solve this problem is by using the equation obtained by equating (1) and (2).{\text{(1) and (2)}}{\text{.}}
⇒−nFEcello=−2.303RTlog⁡K\Rightarrow - nFE_{cell}^{\text{o}} = - 2.303RT\log K
From this, we can isolate the equilibrium constant term to one side:
log⁡K=−nFEcello−2.303RT\log K = \dfrac{{ - nFE_{cell}^{\text{o}}}}{{ - 2.303RT}}
On substituting values for constants F, R and T, we get:
log⁡K=nEcello0.0591\log K = \dfrac{{nE_{cell}^{\text{o}}}}{{0.0591}} at 298K.298K.