# What is the cofficient of Cr^2+?

Apr 11, 2018

$2 C {r}^{2 +}$

#### Explanation:

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

In this case:
$C {r}^{2 +} \left(a q\right) \to C {r}^{3 +} \left(a q\right)$
Is the oxidation
and
$S {O}_{4}^{2 -} \left(a q\right) \to {H}_{2} S {O}_{3} \left(a q\right)$
is the reduction
Start by balancing the half equations for oxygen by adding water:

$S {O}_{4}^{2 -} \left(a q\right) \to {H}_{2} S {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$
(Only the reduction includes oxygen)

Now balance hydrogen by adding protons:

$4 {H}^{+} \left(a q\right) + S {O}_{4}^{2 -} \left(a q\right) \to {H}_{2} S {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$
(again, only the reduction involves hydrogen)

Now balance each half-equation for charge by adding electrons to the more positive side:
$C {r}^{2 +} \to C {r}^{3 +} + {e}^{-}$

$4 {H}^{+} + S {O}_{4}^{2 -} \left(a q\right) + 2 {e}^{-} \to {H}_{2} S {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$

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:

$2 C {r}^{2 +} \to 2 C {r}^{3 +} + 2 {e}^{-}$

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)

$4 {H}^{+} \left(a q\right) + S {O}_{4}^{2 -} \left(a q\right) + 2 C {r}^{2 +} \left(a q\right) \to 2 C {r}^{3 +} \left(a q\right) + {H}_{2} S {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$

Now the equation is balanced and we can see that the coefficient to $C {r}^{2 +}$ is 2.