JEE Main 2026 · 8 Apr · Shift 2 · Chemistry Q56

ChemistrySingle correct+4 / -1Question 56 of 75

Given at 298 K : $E^\ominus_{\ce{Fe^2+}/\ce{Fe}}=X$ Volt, $E^\ominus_{\ce{Fe^3+}/\ce{Fe}}=Y$ Volt

The $E^\ominus_{\ce{Fe^3+}/\ce{Fe^2+}}$ in Volt at 298 K is given by :

2X − 3Y

3Y − 2X

3Y + 2X

Y + X

Answer and worked solution

Official answer: B

Free energies add. $\ce{Fe^3+ + e- -> Fe^2+}$ is $(\ce{Fe^3+ -> Fe})$ minus $(\ce{Fe^2+ -> Fe})$ :
$-1\cdot FE=-3FY+2FX\ \Rightarrow\ E=3Y-2X.$

(B)

Topics:Gibbs energy and EMF

JEE Main 2026 · 8 Apr · Shift 2, Chemistry question 56