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Electrochemistry question

2022 · 25 Jun · Shift 1 · Q20
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  5. /2022 · 25 Jun · Shift 1 · Q20

Electrochemistry question

2022 · 25 Jun · Shift 1 · Q20

JEE MainChemistryElectrochemistryNumerical+4 / −1
In a cell, the following reactions take place Fe2+→Fe3++e−EFe3+/Fe2+o=0.77 V2I−→I2+2e−EI2/I−o=0.54 V\begin{matrix} {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} & {E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.77\,V} \\ {2{I^ - } \to {I_2} + 2{e^ - }} & {E_{{I_2}/{I^ - }}^o = 0.54\,V} \\ \end{matrix}Fe2+→Fe3++e−2I−→I2​+2e−​EFe3+/Fe2+o​=0.77VEI2​/I−o​=0.54V​ The standard electrode potential for the spontaneous reaction in the cell is x ×\times× 10 −-− 2 V 298 K. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest Integer)
Numerical answer
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