Given :
`Hg_(2)^(2+) +2 e ^(-)rightarrow 2Hg`, `E^(@) = 0.789 V` and `Hg^(2+) + 2e^(-) rightarrow Hg`, `E^(@) = 0.854V`
Calculate the equilibrium consant for `Hg_(2)^(2+) rightarrow Hg + Hg^(2+)`.

Given : Hg_(2)^(2+) rightarrow 2Hg , E^(@) = 0.789 V and Hg^(2+) + 2e^(-) rightarrow Hg , E^(@) = 0.854V Calculate the equilibrium consant for Hg_(2)^(2+) rightarrow Hg + Hg^(2+) .

Standard potentials (E°) for some half-reactions are given below: Sn^(4+) + 2e rightarrow Sn^(2+) , E^(@) = + 0.15V 2Hg^(2+) + 2e rightarrow Hg_(2)^(2+) , E^(@) = 0.92 V pbo_(2) + 4H^(+) + 2e rightarrow pb^(2+)+ 2H_(2)O , E^(@) = + 1.45 V Based on the above , Which one of the following statement is correct?

The standard oxidation potential, E^(@) , for the reactions are given as: Zn rightarrow Zn^(2+) + 2e^(-) , E^(@) = +0.76V Fe rightarrow Fe^(2+) + 2e^(-) , E^(@) = +0.41V The emf for the cell : Fe^(2+) + Zn rightarrow Zn^(2+) + Fe

In the reaction, 2HgOrarr2Hg+O_(2),Hg^(2+) act as

For the reduction of NO_(3)^(-) ion in an aqueous solution , E^(0) is + 0.96 V , the values of E^(0) for some metal ions ar given below : i) V_((aq))^(+2) + 2e^(-) to V, E^(0) = 1.19V ii) Fe_((aq))^(+3) + 3e^(-) to Fe , E^(0) = -0.04 V iii) Au_((aq))^(+3) + 3e^(-) to Au , E^(0) = + 1.40 V iv) Hg_((aq))^(+2) + 2e^(-) to Hg , E^(0) = +0.86 V The pair(s) of metals that is/are oxidizd by NO_3^(-) in aqueous solution is /are

Hg_(2)CI_(2) is produced by the electrolytic reduction of Hg^(2+) ion in presence of CI^(-) ion is 2Hg^(2+) +2CI^(-) +2e^(Theta) rarr Hg_(2)CI_(2) . Calculate the current required to have a rate production of 44g per hour of Hg_(2)CI_(2) . [Atomic weight of Hg = 200.6] :-