MnO\(_{4(aq)}^{-1}\) + 8H\(^+ _{(aq)}\) + Y → Mn\(^{2+} _{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
Y in the equation above represents
2e-
3e-
5e-
7e-
Explanation
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Discussions (21)
the equation given isnt correct
it supposed to be
MnO4^-(aq) + 8H+(aq) + Y → Mn2+(aq) + 4H2O(l)
there's a minus sign in MnO4^-
hence
from the reactant side we get
1-8 = 7
den from the product side we get
2
7-2 = 5e//
the answer is correct simple redox equation to balance it without long process
MnO4
we know that O has an oxidation number of -2
Y=-2*4= -6
but because of the O already on the right hand side of the equation i means O was oxidized by loosing an e-
therefore Mn to balance the original MnO4 will take an oxidation number of +5 which is the 5e-
The correct answer is 6e- and is not listed
MnO4(aq) + 8H+(aq) + Y → Mn2+(aq) + 4H2O(l)
first of all we are to balance the atoms.
the atoms are already balanced so we move to balance the charges
on the left the net charges are: +8 and on the right the net charges are +2
. to balance charges on both sides of the equation, Add 6 electrons to neutralize the 6 excess positive charges on LHS.
This gives the balanced equation below:
MnO4(aq) + 8H+(aq) + 6e- → Mn2+(aq) + 4H2O(l)
therefore the answer is 6e-
long calculation but on balancing electron add 5 electrons to the reactants having 8+ already, so we get both product and reactants balance! this is simple redox
I got confused assuming i did not enter here i will be more confused
Pls My school correct your question there is a -1 ion on MnO4 if note people wikk get confused
Thank you!
