Consider the following redox reaction : \(\mathrm{MnO}_{4}^{-}+\mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) The standard reduction potentials are given as below ( \(\mathrm{E}_{\text {red }}^{\circ}\) ) \(\mathrm{E}_{\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{CO}_{2} / \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}}^{\circ}=-0.49 \mathrm{~V}\) If the equilibrium constant of the above reaction is given as \(\mathrm{K}_{\mathrm{eq}}=10^{\mathrm{x}}\), then the value of \(\mathrm{x}=\) ____ (nearest integer)

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Redox Reactions and Electrochemistry

Consider the following redox reaction :

MnO_{4}^{-} + H^{+} + H_{2} C_{2} O_{4} ⇌ Mn^{2 +} + H_{2} O + CO_{2}

The standard reduction potentials are given as below (

E_{red}^{\circ}

)

E_{MnO_{4}^{-} / Mn^{2 +}}^{\circ} = + 1.51 V

E_{CO_{2} / H_{2} C_{2} O_{4}}^{\circ} = - 0.49 V

If the equilibrium constant of the above reaction is given as

K_{eq} = 1 0^{x}

, then the value of

x =

____ (nearest integer)

Answer: 339

For

2 M n O_{4}^{-} + 5 H_{2} C_{2} O_{4} + 6 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

M n O_{4}^{-}

is reduced at the cathode (

E^{\circ} = + 1.51 V

) and

H_{2} C_{2} O_{4}

is oxidised at the anode (

E_{o x}^{\circ} = + 0.49 V

E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} + E_{o x, an o d e}^{\circ} = 1.51 + 0.49 = 2.00 V

Each

M n O_{4}^{-}

gains 5 electrons and two

M n O_{4}^{-}

ions react, so

n = 10

. At equilibrium,

E_{ce l l}^{\circ} = \frac{0.059}{n} lo g K

2.00 = \frac{0.059}{10} lo g K \Rightarrow lo g K = \frac{20}{0.059} = 338.98

Rounded to the nearest integer,

x = 339

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer