Question

The standard cell potential for the reaction
MnO4- + 5Fe+2 + 8H+ -------> Mn2+ + 5Fe+3 + 4H2O
is 0.59V at 298K. Find the equilibrium constant.

Answer 1

${\underline{\sf{Given}}}$

Reaction :

$\sf \:MnO_{4}+5Fe{}^{+2}+8H{}^{+} \rightarrow Mn{}^{+2}</p><p>+5Fe{}^{+3}+4H_{2}O$

• Standard cell potential for this reaction is 0.59 V at 298 k

${\underline{\sf{To\:Find}}}$

Equlilbrum constant of the given reaction .

${\underline{\sf{Theory }}}$

•Nernst Equation :

$\sf \: E_{cell}=E°_{cell}+\dfrac{-2.303RT}{nF}\log \dfrac{[Prduct]}{[Reactant]}$

At Equlilbrum ,$\sf \: E_{cell}=0$

Therefore ,the Nernst Equation is modified as

$\sf0=E°_{cell}+\dfrac{-2.303RT}{nF}log \dfrac{[Prduct] _{equilibrium}}{[Reactant] _{equilibrium}}$

$\sf \implies \: E°_{cell} = \dfrac{2.303RT}{nF}\log(k_{c})$

Where ,Kc= equilibrium constant at 298K

${\underline{\sf{Solution}}}$

Given Reaction :

$\sf \:MnO_{4}+5Fe{}^{+2}+8H{}^{+} \rightarrow Mn{}^{+2}</p><p>+5Fe{}^{+3}+4H_{2}O$

$\rule{200}2$

Reactions :

$\sf \: Anode: Mn{}^{+7}O_{4} \rightarrow \: Mn{}^{+2}+5e{}^{-1}$

$\sf \: Cathode :5Fe{}^{+2}+5e{}^{-1}\rightarrow </p><p>5Fe{}^{+3}</p><p>$

Therefore , n = 5

We know that

$\sf E°_{cell} = \dfrac{2.303RT}{nF}\log(k_{c})$

$\sf \implies E°_{cell} = \dfrac{0.0591}{5}\log(k_{c})$

Given : Standard potential of reaction = 0.59 V

$\sf \implies \:0.59 = \dfrac{0.0591}{5}\log(k_{c})$

$\sf \implies \log(k_{c}) = \dfrac{0.59 \times 5}{0.0591}$

$\sf \implies \log(k_{c}) \approx50$

$\sf \implies k_{c} = 10 {}^{50}$

It is the required solution!

Answer 2

Given :-

$\sf{ {MnO_4}^{-} + {5Fe}^{+2} + 8H \longrightarrow Mn_2 + {5Fe}^{+3} + 4H_2O}$

→ The standard cell potential for equation is 0.59 V at 298K.

To Find :-

Equilibrium constant of equation .

${\underline{\sf{Theory }}}$

At Equlilbrum ,$\sf \: E_{cell}=0$

$\sf \implies \: E°_{cell} = \dfrac{2.303RT}{nF}\log(k_{c})$

Where ,Kc= equilibrium constant at 298K

solution :

Given Reaction :

$\sf \:MnO_{4}+5Fe{}^{+2}+8H{}^{+} \rightarrow Mn{}^{+2}</p><p>+5Fe{}^{+3}+4H_{2}O$

Here →

$\sf \:MnO_{4}$

x+(-8)= -1

x =7

Thus oxidation state of, Mn in MnO4 is +7

On observing the reaction

$\sf \:MnO_{4}+5Fe{}^{+2}+8H{}^{+} \rightarrow Mn{}^{+2}$

Therefore , n = 5

we know that

$\sf E°_{cell} = \dfrac{2.303RT}{nF}\log(k_{c})$

$\sf \implies E°_{cell} = \dfrac{0.0591}{5}\log(k_{c})$

Given : Standard potential of reaction = 0.59 V

$\sf \implies \:0.59 = \dfrac{0.0591}{5}\log(k_{c})$

$\sf \implies \log(k_{c}) = \dfrac{0.59 \times 5}{0.0591}$

$\sf \implies \log( K_{c} ) = 50$

$\sf{This\: implies\: k_c = antilog (50) }$

$\sf{\implies 1 \times {10}^{50}}$