Consider the cell Pt ( s ) H 2 (   s ) ( latm ) H + aq , H + = 1 | | Fe 3 + ( aq ) , Fe 2 + ( aq )…

Consider the cell Pt(s)H2( s)(latm)H+aq,H+=1||Fe3+(aq),Fe2+(aq)Pt(s)

Given: EFe3+/Fe2+°=0.771 V and EH+/12H2°=0 V, T=298 K

If the potential of the cell is 0.712 V the ratio of concentration of Fe2+ to Fe3+ is
(Nearest integer)

Solution

Cell reaction which occurs:
12H2( g)+Fe3+ (aq.) H+(aq)+Fe2+ (aq.) 
Using Nernst' equation:

E=Eo-0.0591logFe2+Fe3+

0.712=(0.771-0)-0.0591logFe2+Fe3+

logFe2+Fe3+=(0.771-0712)0.059=1

Fe2+Fe3+=10

Asked in: JEE Main 2023 (25 Jan Shift 1)

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