The obseved emf of the cell `Pt//H_(2)(g, 1" atm")| H^(+)(3xx10^(-4) M)"||" H^(+)(M_(1)) //H_(2)(g, 1 " atm")//Pt` at 298 K is 0.154 V. Calculate the

Question

The obseved emf of the cell `Pt//H_(2)(g, 1" atm")| H^(+)(3xx10^(-4) M)"||" H^(+)(M_(1)) //H_(2)(g, 1 " atm")//Pt` at 298 K is 0.154 V. Calculate the

The obseved emf of the cell
`Pt//H_(2)(g, 1" atm")| H^(+)(3xx10^(-4) M)"||" H^(+)(M_(1)) //H_(2)(g, 1 " atm")//Pt`
at 298 K is 0.154 V. Calculate the value of `M_(1)`.

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`(1)/(2)H_(2)(g, 1 " atm")rarr H^(+)(3xx10^(-4)M)+e^(-)" "("At cathode")`
`(H^(+)(M_(1))+e^(-)rarr (1)/(2)H_(2)(g, 1 " atm")" " ("At cathode"))/(H^(+)(M_(1))rarrH^(+)(3xx10^(-4) M)" "("Cell reaction")`
According to Nernst equation, emf of cell is :
`E_(cell)=(E_(H^(+)//H_(2))^(@)-E_(H^(+)//H_(2))^(@))-(0.0591)/(1)"log"((3xx10^(-4)))/((M_(1))`
`E_(cell)=0.154 V, E_(H^(+)//H_(2))^(@)=0`
` :. 0.154=0-0.0591 " log"(3xx10^(-4))/(M_(1))`
`0.154=0.0591 " log"(M_(1))/(3xx10^(-4))`
`"log"(M_(1))/(3xx10^(-4))=(0.154)/(0.0591)=2.6057`
`logM_(1)-log(3xx10^(-4))=2.6057`
log`M_(1)`-(log 3-4 log 10)=2.6057
log`M_(1)`+(4 - log 3)=2.6057
log `M_(1)`+(4-0.4771)=2.6057
log`M_(1)`+3.5229=2.6057
log`M_(1)`=2.6057-3.5229=-0.9172
`log M_(1) =bar(1).0828 or M_(1)=antilog bar(1).0828=0.1210 M`