Question

Consider the following redox reaction :

\(Mn O^-_4 + H ^+ + H_2 C_2 O_4 \rightleftharpoons Mn^{2+} + H_2 O + CO _2\)

The standard reduction potentials are given as below \((E ^\circ _ {red})\)

\(E ^\circ _ {MnO^-_4 / Mn ^{2+}} =+1.5 1V\)

\(E^\circ _{CO_2/H_2C_2O_4} = -0.49 V\)

If the equilibrium constant of the above reaction is given as \(K_{eq}\) = \(10^x\) , then the value of x = _______ (nearest integer)

Answer 1

Correct answer is : 338 or 339

\(Cell\,Rx ^n; MnO^-_4 +H_2C_2 O_4 \rightarrow Mn^{2+}+ CO_2\)

\(E^{circ}_{cell}=E ^\circ _{op }\) of anode \(+E ^\circ _ {RP}\) of cathode

= 0.49 + 1.51 = 2.00V 

At equilibrium

\(E_{cell}=0,\)

\(E ^\circ _{cell}= \frac {0.059}{n}log \,K\)

(As per NCERT \(\frac {RT}{F} = 0.059\) But \(\frac {RT}{F} = 0 .0 591\) can also be taken.)

\(2 = \frac {0.059}{10} log K\)

logK = 338.98