In a cell, the following reactions take place
${Fe}^{2 +} \to {Fe}^{3 +} + e^{-} {E^{o}}_{{Fe}^{3 +} / {Fe}^{2 +}} = 0.77 V$
$2 I^{-} \to I_{2} + 2 e^{-} E_{I_{2} / I^{-}}^{0} = 0.54 V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{- 2} V at 298 K$ . The value of $x$ is - (Nearest Integer)

Question

In a cell, the following reactions take place

${Fe}^{2 +} \to {Fe}^{3 +} + e^{-} {E^{o}}_{{Fe}^{3 +} / {Fe}^{2 +}} = 0.77 V$

$2 I^{-} \to I_{2} + 2 e^{-} E_{I_{2} / I^{-}}^{0} = 0.54 V$

The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{- 2} V at 298 K$ . The value of $x$ is - (Nearest Integer)

Aicharya · Accepted Answer

The electrode which has more reduction potential will get reduced in spontaneous reaction in the cell. So, spontaneous reaction will be

$2 {Fe}^{3 +} (aq) + 2 I^{-} (aq) \to 2 {Fe}^{2 +} (aq) + I_{2}$

Iron will act as cathode and Iodine will act as anode.

${E^{0}}_{cell} = {({E^{0}}_{RP})}_{C} - {({E^{0}}_{RP})}_{A}$

$= 0.77 - 0.54$

$= 0.23 V$

$= 23 \times 10^{- 2} V$

In a cell, the following reactions take place Fe 2 + → Fe 3 + + e -    E o Fe 3 + / Fe 2 + =…