In a cell, the following reactions take place Fe^(2+)→Fe^(3+)+e^(-) E^oFe^(3+)/Fe^(2+)=0.77 V 2I^(-)→I_2+2e^(-) EI_2/I^(-)0=0.54 V The standard…

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)

MARKS Content Team · Accepted Answer

Correct answer: 23. The electrode which has more reduction potential will get reduced in spontaneous reaction in the cell. So, spontaneous reaction will be 2Fe^(3+)aq+2I^(-)aq→2Fe^(2+)aq+I_2 Iron will act as cathode and Iodine will act as anode. E^0cell=E^0RP_C-E^0RP_A =0.77-0.54 =0.23 V =23×10^(-2) V

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In a cell, the following reactions take placeFe2+→Fe3++e- EoFe3+/Fe2+=0.77 V2I-→I2+2e- EI2/I-0=0.54 VThe standard electrode potential for the spontaneous reaction in the cell is x×10-2 V at 298 K. The value of x is - (Nearest Integer)

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