Question

b.b/ pts When the following redox reaction is balanced (assume acidic solution), how many moles of water appear, and on what side: MnO4 + Cl --> Mn2+ + Cl2 4. right side 4, left side 8. left side 8, right side

Answer 1

Write the given redox reaction as follows Mn2Cl MnO C Write the reacti ons as follows right si de and 1 Cl on left side so multiply Cl with 2 There are two 2 Cl atoms on to balance the Cl atoms 2C1_ Cl2 Now the charge right si de is 0, so add 2e on left side is - 2 and the charge on right side to bal ance the equation on 2C1 () Ci22e Mn2 2+ MnO4 There are 4 O atoms on left side, so add 4H,O on right side to balance C atoms Mn2 +4H20 MnO right side, so add 8H on left si de to balance the H atoms There are 8 H atoms on 2+ Mn MnO 8H 4H2O right side is +2, so add 5e on left si de to bal ance on left side is+7 and on charge the equation Mn24H2O MnO4 8H +5e (2) multiply equati on (1) with 5 and multiply equation (2) with 2 and add them as follows 5x(1) 5Cl210e 10CI 2Mn 2x(21): 2MnO +16H* +10 e 8H2O + 2Mn2 +5C1 +8H2O 2MnO 10C116H The balanced equation is 2Mn 2+ + 2MnO10CI 16H -5C12 8H2O + right si de There are 8 water molecules on Therefore, the required solution is 8, right si de