Calculate the emf of the following cell at `25^(@)C`. `Zn|Zn^(2+)`(0.001 M) || `H^(+)(0.01 M) | H_(2)`(1 bar) | Pt(s). Given that `E_((Zn^(2+)//Zn))^(@)=-0.76V, E_((H^(+)//H_(2)))^(@)=0.00" V "`

Calculate the emf of the following cell at `25^(@)C`. `Zn|Zn^(2+)`(0.0

1.	Calculate the emf of the following cell at `25^(@)C`. `Zn\|Zn^(2+)`(0.001 M) \|\| `H^(+)(0.01 M) \| H_(2)`(1 bar) \| Pt(s). Given that `E_((Zn^(2+)//Zn))^(@)=-0.76V, E_((H^(+)//H_(2)))^(@)=0.00" V "`
Answer» Correct Answer - 0.7305V According to Nernst equation : `E_(cell)=E_(cell)^(@)-(0.0591V)/(n)"log"([Zn^(2+)])/([H^(+)]^(2))` `E_(cell)=(0.76" V")-((0.0591" V"))/(2)"log"(10^(--3))/((10^(-2))^(2))` `=0.76" V "-(0.02955" V ")" log "10=(0.76-0.2955)" V "=0.7305" V "`

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Calculate the emf of the following cell at `25^(@)C`. `Zn|Zn^(2+)`(0.001 M) || `H^(+)(0.01 M) | H_(2)`(1 bar) | Pt(s). Given that `E_((Zn^(2+)//Zn))^(@)=-0.76V, E_((H^(+)//H_(2)))^(@)=0.00" V "`

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