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Dear experts, Could you please answer these questions

1-

2-

3-

4-

5-

Thank you

For the following reaction assign oxidation states to each element on each side of the equation. Which element is oxidized? Which element is reduced? The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.17-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCI(aq) and passed over a reducing agent so that all the antimony is in the form Sb3+(aq). The Sb3+(aq) is completely oxidized by 25.0 mL of a 0.140 M aqueous solution of KBrQ3(aq). The unbalanced equation for the reaction is (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Calculate the E degree cell for the following equation. Use these standard potentials. A galvanic (voltaic) cell consists of an electrode composed of nickel in a 1.0 M nickel(ll) ion solution and another electrode composed of copper in a 1.0 M copper(ll) ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 degree C. Standard reduction potentials can be found here. Calculate the standard free-energy change for the following reaction at 25 degree C. Standard reduction potentials can be found here.

Explanation / Answer

1.

oxidation states

Reactant Product

K 1 1

Cl +5 -1

O -2 0

O is oxidized

Cl is reduced

" Chlorine " in KClO3 is reduced from Oxidation No. = ( +5 ) to " Chlorine " in KCl _ Oxidation No. = ( -1 )
KClO3 __ [ K ion = +1 ; 3 O = ( 3 x -2 ) ; Cl = +5 ; Sum = 0 ]
KCl __ [ K ion = +1 ; Cl ion = -1 ; Sum = 0 ]

" Oxygen " in KClO3 is oxidised from Oxidation No. = ( -2 ) to " Oxygen " in O2 _ Oxidation No. = ( 0 )
KClO3 __ [ K ion = +1 ; 3 O = ( 3 x -2 ) ; Cl = +5 ; Sum = 0 ]
O2 (g) __ [ O = 0 ; By Definition ]

Oxidation No. of all other atoms remain unchanged.

2.

First we have to balance the equation. We do that by balancing each half-reaction separately, then combining them.

BrO3- ==> Br- . . .Add 3 H2O to the right side to balance O.
BrO3- ==> Br- + 3H2O . . .add 6H+ (acidic solution) to the left side to balance H.
BrO3- + 6H+ ==> Br- + 3H2O . . .add 6e- to the left side to balance the charge.
BrO3- + 6H+ + 6e- ==> Br- + 3H2O

Sb3+ ==> Sb5+ . . .Add 2e- to the right side to balance the charge.
Sb3+ ==> Sb5+ + 2e-

Multiply the Sb equation by 3 to give 6e- on the right side. Then add the new equation to the BrO3- equation; the 6e- will cancel.

.3Sb3+ ==> 3Sb5+ + 6e-
+BrO3- + 6H+ + 6e- ==> Br- + 3H2O
===================================
3Sb3+ + BrO3- + 6H+ ==> 3Sb5+ + Br- + 3H2O


moles BrO3- added = M BrO3- x L BrO3- = (0.140)(0.0250) = 0.00350 moles BrO3-

The balanced equation tells us that 1 mole of BrO3- reacts with 3 moles of Sb3+.

0.00350 moles BrO3- x (3 mole Sb3+ / 1 mole BrO3-) = 0.0105 moles Sb3+

From the periodic table, the molar mass of Sb (or Sb3+; the 3 missing electrons have very little effect on the mass ) = 121.8 g.

0.0105 moles Sb x (121.8 g Sb / 1 mole Sb) = 1.28 g Sb

%Sb = (g Sb / g ore) x 100 = (1.28 / 6.17) x 100 = 20.7 %Sb

3.

Cu(s) + Ag+(aq) --> Cu+ (aq) + Ag(s)
Ecell____0.28______ V.

The two half reactions are:
Ag+(aq) + e- ? Ag(s) reduction E

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