Calculate the equilibrium constant for the reaction, `2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_3^-` . The standard reduction potentials in acidic conditions are 0.77 and 0.54 V respectively for `Fe^(3+)//Fe^(2+)` and `I_3^(-) //I^-` couples.

Calculate the euilibrium constant for the reaction, 2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-) . The standard reduction potential in acidic conditions are 0.77 V and 0.54 V respectivelu for Fe^(3+)//Fe^(2+) and I_(3)^(-)//I^(-) couples.

Write the half reaction for the reaction 2Fe^(+3) + 2I^(-) to 2Fe^(+2) + I_(2)

Write the equilibrium constant expression for the following reactions : Fe^(3+) (aq) + SCN^(-) (aq) hArr FeSCN^(2+) (aq)

Calculate the n-factor of reactants in the given reactions. (d) Fe(NO_(3))_(3)rarr Fe^(2+)+NO

The equilibrium constant of the following redox rection at 298 K is 1 xx 10^(8) 2Fe^(3+) (aq.) +2I^(-) (aq.) hArr 2Fe^(2+)(aq.) +I_(2) (s) If the standard reducing potential of iodine becoming iodide is +0.54 V. what is the standard reduction potential of Fe^(3+)//Fe^(2+) ?

Calculate the equilibrium constant for the reaction Fe(s)+Cd2+(aq)⇔Fe2+(aq)+Cd(s) (Given : E∘Cd2+∣Cd=0.40V,E∘Fe2+∣Fe=−0.44V).

Given that the standard reduction potentials E ^(0) of Fe^(+2)|Feis 0.26V and Fe ^(+3) |Fe is 0.76V respectively. The E ^(@)of Fe ^(+2)|Fe^(+3) is:

Determine the equilibrium constant of the reaction at 298 K, 2Fe^(3+) +Sn^(2+) hArr 2Fe^(2+) +Sn^(4+) From the obtained value of the equilibrium constant, predict whether Sn^(2+) ions can reduce Fe^(3+) to Fe^(2+) quantitatively or not. Given: E_(Fe^(3+),Fe^(2+)//Pt)=0.771V.E_(Sn^(4+)//Sn^(2+)//Pt)^(Theta)=0.150V . [R=8.314JK^(-1)"mol"^(-1)]

Find the standard electrode potential of I_(2)|2I^(c-) if the equilibrium constant for the reaction I_(2)+I^(c-) rarr_'(I_3)^(c-) is 703. The standard electrode potential of I_(3)^(c-)|3I^(c-) is 0.5355V . Also give the electrode reaction.

For the reaction Fe_2N(s)+(3)/(2)H_2(g)=2Fe(s)+NH_3(g)