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What does the negative sign in the expre...

What does the negative sign in the expression `E_(Zn^(2+)//Zn)^(@)=-0.76V` mean?

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Greater the negative reactivity of standard reduction potential of metal greater is its reactivity. It means that Zn is more reactive than hydrogen. When zinc electrode will be connected to standard hydrogen electrode Zn will get oxidised and `H^(+)` will get reduced. thus, zinc electrode will be the anode of the cell and hydrogen electrode will be the cathode of the cell.
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The emf of a cell corresponding to the reaction Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1 atm is 0.28 volt at 25^(@)C . Calculate the pH of the solution at the hydrogen electrode. E_(Zn^(2+)//Zn)^(@) =- 0.76 volt and E_(H^(+)//H_(2))^(@) = 0

Can a solution of 1 M ZnSO_(4) be stored in a vessel made of copper ? Given that E_(Zn^(+2)//Zn)^(@) =-0.76V and E_(Cu^(+2)//Cu)^(@)=0.34 V

Why zinc coating protects iron more effectively than tin coating ? Given E_(Zn^(2+)//Zn)^(Theta)= -0.76V , E_(Sn^(2+)//Sn)^(Theta)= -0.14V, E_(Fe^(2+)//Fe)^(Theta)= -0.44V

One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration . Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO_(3))_(2) . A voltage of 1.48 V is measured for this cell . Use this information to calculate the concentration of silver nitrate solution used. (E_(Zn^(2+)|Zn)^(@) = -0.76V , E_(Ag^(+)|Ag)^(@) = + 0.80V ).

What does the standard electrode potential of a metal being negative (E_(Zn^(2+)//Zn)^(Theta)= -0.7632) indicate ?

Explain the following : Zinc displaces hydrogen from acid solution. E_(Zn^(2+)//Zn)^(Theta)= -0.76V

Which of the following will act as cathode when connected to standard hydrogen electrode which has E^(@) value given as zero ? (i) Zn^(2+)//Zn, E^(@)=-0.76V (ii) Cu^(2+)//Cu, E^(@)=+0.34 V (iii) Al^(3+)//Al, E^(@)=-1.66 V (iv) Hg^(2+)//Hg, E^(@)=+0.885 V (A)(i) and (ii) (B)(ii) and (iv) (C)(i) and (iii) (D)(i), (ii), (iii) and (iv)

For the galvanic cell Zn(s)[Zn^2+ (aq)||Ag^+(aq) Ag(s)] , G° of the reaction in joule will be E°_ (Zn^(2+) / Zn )= - 0.76V , E°_( Ag_+/Ag )= 0.80 V

Calculate the equilibrium constant for the reaction at 298K. Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.76V and E_(Cu^(2+)//Cu)^(@) = +0.34 V

Calculate the equilibrium constant for the reaction at 298K. Zn(s) +Cu^(2+)(aq) hArr Zn^(2+)(aq) +Cu(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.76V and E_(Cu^(2+)//Cu)^(@) = +0.34 V

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