In an electrochemical reaction of lead, at standard temperature, if $E^0_{(\ce{Pb^{2+}/Pb})} = m$ Volt and $E^0_{(\ce{Pb^{4+}/Pb})} = n$ Volt, then the value of $E^0_{(\ce{Pb^{2+}/Pb^{4+}})}$ is given by $m - xn$. The value of $x$ is $\_\_\_\_$. (Nearest integer)

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Redox Reactions and Electrochemistry

In an electrochemical reaction of lead, at standard temperature, if

E_{(Pb X^{2 +} / Pb)}^{0} = m

Volt and

E_{(Pb X^{4 +} / Pb)}^{0} = n

Volt, then the value of

E_{(Pb X^{2 +} / Pb X^{4 +})}^{0}

is given by

m - x n

. The value of

x

____

.
(Nearest integer)

Answer: 2

Standard potentials are not additive, but free energies are, through

Δ G^{\circ} = - n F E^{\circ}

. For

Pb X^{2 +} + 2 e X^{-} Pb

Δ G_{1}^{\circ} = - 2 F m

; for

Pb X^{4 +} + 4 e X^{-} Pb

Δ G_{2}^{\circ} = - 4 F n

. The required change

Pb X^{2 +} Pb X^{4 +} + 2 e X^{-}

Δ G^{\circ} = Δ G_{1}^{\circ} - Δ G_{2}^{\circ} = - 2 F m + 4 F n

, and with

Δ G^{\circ} = - 2 F E^{\circ}

this gives

E^{\circ} = m - 2 n

. Comparing with

m - x n

, the value of x is 2.

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Integer answer