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what happenes to the reduction potential of Pt | H2 (Patm) H + (1M) at 298 K when P increases from 1 atm to 10 atm what happenes to the reduction potential of Pt | H2 (Patm) H+(1M) at 298 K when P increases from 1 atm to 10 atm
it changes from 0 to 0.03 am i correct....?if yes i will say u how
YES SIR CAN TELL ME HOW
apply nernst equationH2gives 2H+ +2e-so here n factor is 2 Ecell=E0 -((0.06)/n)log(1/10)=0.03APPROVE if useful and ask if u have any doubts please refer cengage for iit jee (chem)if u have diffucalty in physical
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