Balanced redox reaction for Sn(NO3)2(aq) and acidifed KMnO4 (aq)
First, let's write out the reactants and products of each compound:
Sn(NO3)2(aq) → Sn2+(aq) + 2NO3-(aq)
KMnO4(aq) → K+(aq) + MnO4-(aq)
Next, let's write out the half-reactions for each compound:
Sn2+(aq) → Sn4+(aq) + 2e-
MnO4-(aq) → Mn2+(aq)
Now, we need to balance the number of electrons in each half-reaction:
2Sn2+(aq) → 2Sn4+(aq) + 4e-
5MnO4-(aq) + 8H+(aq) + 5e- → 5Mn2+(aq) + 4H2O(l)
To balance the overall reaction, we need to multiply the equations by appropriate coefficients so that the number of electrons in each half-reaction cancels out:
10Sn2+(aq) + 5MnO4-(aq) + 16H+(aq) → 10Sn4+(aq) + 5Mn2+(aq) + 8H2O(l)
Therefore, the balanced redox reaction for Sn(NO3)2(aq) and acidified KMnO4(aq) is:
10Sn2+(aq) + 5MnO4-(aq) + 16H+(aq) → 10Sn4+(aq) + 5Mn2+(aq) + 8H2O(l)