Electrochemistry Find
`Sn^(4+) +4e^-rarr Sn ` `E^o` = 0.010eV
`Sn^(2+) +2e^- rarr Sn ` `E^o` = -0.16eV
`Sn^(4+) +2e^-rarr Sn^(2+) ` `E^o` = ?

Pb^(+2) + 2e^(-) rarr Pb(s), E^(@) = -0.13V Sn^(+2) + 2e^(-) rarr Sn(s), E^(@) = - 0.16V Ni^(+2) + 2e^(-) rarr Ni(s), E^(@) = -0.25V Cr^(+3) + 3e^(-) rarr Cr(s), E^(@) = -0.74V Based on the above data, the reducing power of Pb, Sn, Ni and Cr is in the order

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?

According to the half-reaction table, Sn^(2+) + 2e^(-) rightarrow Sn E_(@) = -0.14V Mn^(2+) + 2e^(-) rightarrow Mn E^(@) = -1.03V Which species in the better oxidizing agent?

Use the standard reduction potentials: Sn^(2+)(aq) + 2e^(-) rightarrow Sn(s) E^(@) = -0.141V Ag^(+)(aq) + e^(-) rightarrow Ag(s) E^(@) = 0.800V To calcultate E^(@) for the reaction: Sn(s) + 2Ag^(+)(aq) rightarrow Sn^(2+)(aq) + 2Ag(s)

Cu^(2+) + Sn rarr Sn^(2+) + Cu E^@_cu^(2+)/cu = 0.34, E^@_Sn^(2+)/Sn = -0.16 . find deltaG^@

The standard electrode potentials for the reactions, Ag^(+) (a.q) + e^(-) rarr Ag (s) Sn^(2+) (a.q) + 2e^(-) rarr Sn (s) at 25^(@)C are 0.80 volt and -0.14 volt respectively. The emf of the cell Sn|Sn^(2+) (1M)|| Ag^(+) (1M) | Ag is: