2 e(-) + CN(-) --> CNO(-)

2 e(-) + 2 H(+) + CN(-) --> CNO(-) + H2O

2 e(-) + 2 H2O + CN(-) --> CNO(-) + H2O + 2 OH(-)

2 e(-) + H2O + CN(-) --> CNO(-) + 2 OH(-)

MnO4(-) --> MnO2 + 3 e(-)

4 H(+) + MnO4(-) --> MnO2 + 3 e(-) + 2 H2O

4 H2O + MnO4(-) --> MnO2 + 3 e(-) + 2 H2O + 4 OH(-)

2 H2O + MnO4(-) --> MnO2 + 3 e(-) + 4 OH(-)

They are all wrong. You failed to understand the concept behind half-equations.

i show you an example question to illustrate how to do.

Balance MnO

_{4}^{-} + Fe

^{2+} => Mn

^{2+} + Fe

^{3+}Do not forget that electron carries the charge of -1.

We first identify: Fe

^{2+} => Fe

^{3+}For Iron(II) to be oxidised to Iron(III), it must loose 1 electron.

Fe

^{2+} => Fe

^{3+} + e

^{-}Next, we have to balance MnO

_{4}^{-} => Mn

^{2+}since we are dealing with an aqeous redox system, it is valid to assume H+ or OH- or water molecules might participate in the reaction.

we observe that the reduced form of manganese has no oxygen, so H+ must have participated in the half-equation and convert the oxygen to water.

MnO

_{4}^{-} + 8H

^{+} => Mn

^{2+} + 4H

_{2}O

the overall charge on the LHS is 7+ and the overall charge on the RHS is 2+. This means the LHS somehow must acquire -5 charge in order for the equation to balance. This means addition of 5 electrons to the LHS. Remember that reduction is also defined as gaining electrons.

MnO

_{4}^{-} + 8H

^{+} + 5e

^{-} => Mn

^{2+} + 4H

_{2}O

Now, the manganese half-equation is finally balanced, charge-wise and material-wise.

(1) MnO

_{4}^{-} + 8H

^{+} + 5e

^{-} => Mn

^{2+} + 4H

_{2}O

(2) Fe

^{2+} => Fe

^{3+} + e

^{-}We seek to combine both equations. Redox reactions involve the transfer of electrons. This means the number of electron donated by the oxidation half-equation must equal the number of electron accepted by the reduction half-equation.

Equation (1) involves 5 electrons, but equation (2) involves 1 electron. Since the lowest common multiple between 1 and 5 is 5, then we must scale up the equation (2) 5 times.

Finally, we combine Equation (1) + 5 X Equation (2)

ie. LHS of Equation (1) + 5 X LHS of Equation (2) => RHS of Equation (1) + 5 X RHS of Equation (2)

MnO

_{4}^{-} + 8H

^{+} + 5e

^{-} + 5Fe

^{2+} => Mn

^{2+} + 4H

_{2}O + 5Fe

^{3+} +5 e

^{-}Next, eliminate all repeated terms on the LHS and RHS, so we yield the final equation:

MnO

_{4}^{-} + 8H

^{+} + 5Fe

^{2+} => Mn

^{2+} + 4H

_{2}O + 5Fe

^{3+}I hope you understood my workings.