okay i have done this
Mn2+ equi MnO4- + 5e
BiO3- +2e equi Bi 3+
so
Mn2+ + 4H20 equi MnO4- + 5e + 8H+
BiO3- + 2e + 6G equi Bi3+ + 3H20
common factor of 10 so multiplied top reaction by 2 and bottom by 5 and came to final answer of
2Mn2+ + 5BiO3- + 14H+ equi 2MnO4- + 5Bi3+ + 7H20
....i think its right, but what im confused about for having to do the cell notation is which is the oxidation and which is the reducution because i know oxidation occurs in the anode, so can you help me out with which is which I know the Bi gained two electrons so i guess the Bi is reduction but im still confused. please