Pt ( s ) H 2 (   g ) ( 1 bar ) H + ( aq ) ( 1 M ) M 3 + ( aq ) , M + ( aq ) Pt ( s ) The E cell …

Pt(s)H2( g)(1bar)H+(aq)(1M)M3+(aq),M+(aq)Pt(s)

The Ecell  for the given cell is 0.1115 V at 298 K

When M+(aq)M3+(aq)=10a

The value of a is ___________

Given : Eθ=M3+/M+0.2 V

2.303RTF=0.059 V

 

Solution

Overall cell reaction :-

H2( g)+M(aq)3+M(aq)++2H(aq)+

Using Nernst equation:
ECell =ECathode o-Eanode o-0.0592logM+×12M+31

0.1115=0.2-0.0592logM+M+3

3=logM+M+3

a=3

Asked in: JEE Main 2023 (25 Jan Shift 2)

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