`Cu^+ // Cu^2+ , (0.15)`, `Fe^2+//Fe^3+ ,(-0.77)`, `Hg^+ // Hg^+2 (-0.92), 2 Br^(-)`// `Br_2 (-1.07)`.The oxidation potentials are given above, which is the best reducing agent ?

E^@ Fe //Fe^(2+) = 0.44 whereas E^@ Cu // Cu^2+ = -0.32 v Then

Given the standard electrode potentials, K^(+) // K = - 2.93 V, Ag^(2+) //Ag = 0.80 V Hg^(2+) //Hg = 0.79 V , Mg^(2+) //Mg = - 2.37 V, Cr^(3+) //Cr = - 0.74 V Arrange these metals in their increasing order of reducing power.

Between Cr^(2+) and Fe^(2+) which is a stronger reducing, agent in aqueous solution and why?

In the conversion of Br_2 to BrO_3^- ,the oxidation state of bromine changes from :

Which is a stronger reducing agent, Fe^(2+) or Cr^(2+) and why ?

The standard reduction potential at 290K for the following half reactions are, (i) Zn^(2+) +2e rarrZn(s),E^@ =-0.762 V (ii) Cr^(3+) +3e rarr Cr(s), E^@=-0.740 V (iii) 2H^+ 2e rarrH_2(g),E^@=-0.000V (iv) Fe^(3+) +e rarr Fe^2+,E^@=+0.77V Which it the strongest reducing agent :

E^@ Fe^3+ // Fe = -0.036V , E^@ Fe^2+ // Fe = -0.439 V . The value of standard electrode potential for the charge Fe underset(aq) overset(3+) + e rarr Fe underset (aq) overset(2+) will be

Cu (II) sulphate solution is treated separately with KCl and KI In which case, Cu^(2+) be reduced to Cu^+ :