MnO\(_4\)(aq) + Y + 5Fe\(^{2+}\) (aq) \(\mapsto\) Mn\(^{2+}\) (aq) + 5Fe\(^{3+}\) (aq) + 4H\(_2\)O (l).

In the equation above, Y is

SammyKush011

6 years ago

MnO4^– +5Fe2+ +8H+ --> Mn2+ + 5Fe3+ + 4H2O

Total number of charges on both sides 17+, as opposed to your question
LHS 10+ RHS 12+

Toclay

5 years ago

I no get am oo