Question

Acidic solution In acidic solution, the iodate ion can be used to react with a number of metal ions. One such reaction is IO3−(aq)+Sn2+(aq)→I−(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: IO3−(aq)+Sn2+(aq)+ −−−→I−(aq)+Sn4+(aq)+ −−−

Part A-

What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms.

Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Part B-

Potassium permanganate, KMnO4, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In basic solution, the following equation represents the reaction of this ion with a solution containing sodium sulfite: MnO4−(aq)+SO32−(aq)→MnO2(s)+SO42−(aq) Since this reaction takes place in basic solution, H2O(l) and OH−(aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: MnO4−(aq)+SO32−(aq)+     −−−→MnO2(s)+SO42−(aq)+     −−− Part B What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH−(aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes. What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH−(aq) in the blanks where appropriate. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

Answer 1

tayt - H₂11) It snith It suth (92) 191) 1921 192) - Io Isnt2 i I snty 1-79 X -) xat5 - 3 X+3(-2)=-) Xitz 1 oxidation half reaction Reduction half reaction snte ->snity IO I T to balance oxygen atoms | by adding to molecules snt2 - snth IOI 't 340 balance hydrogen atoms by adding the 10 + 6HH - It 340 balance charge by adding electrons snt2 - snt4+ 28t 10 + 6H + tel I'+ 340 multiply with suitable integer to equalise number of electrons DX3 3 snt233nt4 tbet 10 +64th+bel I'+ 3H₂0 snth sht OXI 2 add both 3 snt2 - 3sn't4 te 107 + 64th+bet - I'+ 3H₂o 3 snt2 + 1 g + 6H 35nth + I'+ 340 199) cae) 192) 1922 1921

magitt sore – ning & sq2 t Ho mingi 302 mng 150 m2 on oh! X+46-2)=-1 | X+3(-2)=-2 x+2(-2)=0 | X+41-2)=-2 x=t7 12=+4 x=+4 Txato oxidation half reaction | Reduction half Reaction soe +592 ming" - mną balance oxygen atoms by adding of balance Hidrogen atoms by adding to repeat this process up to balanceing of all Hydrogen and oxygen atoms so 2 + 204 50 2tHo mingit 2H₂0 - mnq+40H balance charge by adding electrons so ² + 2 Offl 30 + Hot 20 mnon + 2H₂O + 3el- mno +4 017' multiply with suitable integer to equalise X2 0x3 350 h + Golf'- 350 + 340+ bel/2mno +44 +6 El-> 2.mnots of add both $ 350 + bont - 350 + 30+ bra 2 mno + 4H₂O toe > Lang+ 8 ot' Zano + 359 ² + 1,0 - 2 ming +352 + 286) electrons. number of