Question

What is the cofficient of Cr^2+?

Answer 1

`2Cr^(2+)`

Explanation:

Start by finding what has been oxidized and what has bee reduced by inspecting the oxidation numbers:

In this case:
`Cr^(2+)(aq) -> Cr^(3+)(aq)`
Is the oxidation
and
`SO_4^(2-)(aq) -> H_2SO_3(aq)`
is the reduction
Start by balancing the half equations for oxygen by adding water:

`SO_4^(2-)(aq) -> H_2SO_3(aq) + H_2O(l)`
(Only the reduction includes oxygen)

Now balance hydrogen by adding protons:

`4H^(+)(aq)+SO_4^(2-)(aq) ->H_2SO_3(aq) + H_2O(l)`
(again, only the reduction involves hydrogen)

Now balance each half-equation for charge by adding electrons to the more positive side:
`Cr^(2+) -> Cr^(3+) + e^-`

`4H^(+)+SO_4^(2-)(aq)+2e^- -> H_2SO_3(aq) + H_2O(l)`

And to equalize the electrons, multiply the whole half equation with the least electrons by an integer to equal the other half-equation in the number of electrons, thereby balancing the electrons on both sides of the equation:

`2Cr^(2+) -> 2Cr^(3+) + 2e^-`

Now combine everything and remove the electrons (as they in equal amounts on both sides they can be canceled in this step - otherwise just simplify as far as possible)

`4H^(+)(aq)+SO_4^(2-)(aq)+2Cr^(2+)(aq) -> 2Cr^(3+)(aq)+H_2SO_3(aq) + H_2O(l)`

Now the equation is balanced and we can see that the coefficient to `Cr^(2+)` is 2.