E deg(Sn^4+/Sn^2+) = 0.15 V and E deg(Fe^3+/Fe^2+) = 0.77 V at 298 K. Consider Sn^2+ + 2Fe^3+ -> Sn^4+ + 2Fe^2+. Take (2.303RT)/(F) = 0.0591 V and F = 96500 C mol^-1. Which statements are correct?

Delta_r G deg = -119.66 kJ mol^-1

If the reaction is written as 1/2 Sn^2+ + Fe^3+ -> 1/2 Sn^4+ + Fe^2+, E deg is unchanged but K becomes 10^10.5

Delta_r G deg = -119.66 kJ mol^-1

If the reaction is written as 1/2 Sn^2+ + Fe^3+ -> 1/2 Sn^4+ + Fe^2+, E deg is unchanged but K becomes 10^10.5

Increasing [Fe^3+] increases the equilibrium constant K

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ec070

$E^{\circ} (Sn X^{4 +} / Sn X^{2 +}) = 0.15 V$ and $E^{\circ} (Fe X^{3 +} / Fe X^{2 +}) = 0.77 V$ at 298 K. Consider $Sn X^{2 +} + 2 Fe X^{3 +} Sn X^{4 +} + 2 Fe X^{2 +}$ . Take $\frac{2.303 RT}{F} = 0.0591 V$ and $F = 96500 C mo l^{- 1}$ . Which statements are correct?

A $lo g_{10} K \approx 21$
B $Δ_{r} G^{\circ} = - 119.66 kJ mo l^{- 1}$
CIf the reaction is written as $\frac{1}{2} Sn X^{2 +} + Fe X^{3 +} \frac{1}{2} Sn X^{4 +} + Fe X^{2 +}$ , $E^{\circ}$ is unchanged but $K$ becomes $1 0^{10.5}$
DIncreasing $[Fe X^{3 +}]$ increases the equilibrium constant $K$

E^{\circ} = 0.77 - 0.15 = 0.62 V

n = 2

lo g K = 2 \times 0.62/0.0591 = 20.98 \approx 21

(A).

Δ_{r} G^{\circ} = - 2 \times 96500 \times 0.62 = - 119660 J = - 119.66 kJ

(B). Halving the equation halves

n

and

Δ_{r} G^{\circ}

, so

E^{\circ}

stays

0.62 V

while

lo g K

halves to

10.5

(C).

K

depends only on temperature, not on concentrations; changing

[Fe X^{3 +}]

shifts the position of equilibrium but not

K

(D wrong).