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Electrochemistry question

2023 · 25 Jan · Shift 1 · Q16
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  5. /2023 · 25 Jan · Shift 1 · Q16

Electrochemistry question

2023 · 25 Jan · Shift 1 · Q16

JEE MainChemistryElectrochemistryNumerical+4 / −1
Consider the cell Pt(s)∣H2(g) (1 atm)∣H+ (aq,[H+]=1)∣∣Fe3+(aq),Fe2+(aq)∣Pt(s)\mathrm{Pt(s)|{H_2}(g)\,(1\,atm)|{H^ + }\,(aq,[{H^ + }] = 1)||F{e^{3 + }}(aq),F{e^{2 + }}(aq)|Pt(s)}Pt(s)∣H2​(g)(1atm)∣H+(aq,[H+]=1)∣∣Fe3+(aq),Fe2+(aq)∣Pt(s) Given EFe3+/Fe2+o=0.771 V\mathrm{E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.771\,V}EFe3+/Fe2+o​=0.771V and EH+/1/2 H2o=0 V, T=298 K\mathrm{E_{{H^ + }/1/2\,{H_2}}^o = 0\,V,\,T = 298\,K}EH+/1/2H2​o​=0V,T=298K If the potential of the cell is 0.712 V, the ratio of concentration of Fe 2+^{2+}2+ to Fe 3+^{3+}3+ is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
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