In a cell, the following reactions take place \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} \mathrm{e}^{-}\) \[ \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\mathrm{o}}=0.77 \mathrm{~V} \] \(2 \mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+2 \mathrm{e}^{-}\) \[ \mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\mathrm{o}}=0.54 \mathrm{~V} \] The standard electrode potential for the spontaneous reaction in the cell is \(x \times 10^{-2} V 298 K\). The value of \(x\) is \(\_\_\_\_\) (Nearest Integer)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Redox Reactions and Electrochemistry

In a cell, the following reactions take place

Fe^{2 +} \to Fe^{3 +} e^{-}

E_{Fe^{3 +} / Fe^{2 +}}^{o} = 0.77 V

2 I^{-} \to I_{2} + 2 e^{-}

E_{I_{2} / I^{-}}^{o} = 0.54 V

The standard electrode potential for the spontaneous reaction in the cell is

x \times 1 0^{- 2} V 298 K

. The value of

x

____

(Nearest Integer)

Answer: 23

For a spontaneous cell reaction, Fe³⁺ (higher reduction potential) acts as cathode and I⁻ acts as anode.

The cell reaction is:

Fe^{3 +} + I^{-} \to Fe^{2 +} + I_{2}

Standard cell potential:

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}

= 0.77 - 0.54 = 0.23 V

= 23 \times 1 0^{- 2} V

Therefore, x = 23.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer