Question

some chemical reactants are listed in the table below. complete the table by filling in the oxidation state of the highlighted atom.

Explanation:

Step1: Oxidation state of O in \(H_3O^+\)

Let the oxidation state of O be \(x\).
For \(H_3O^+\), we know that the oxidation state of H is \(+ 1\).
Using the formula: \(3\times(+1)+x=+1\) (since the overall charge of \(H_3O^+\) is \(+1\))
\(3 + x=+1\)
\(x=+1 - 3=-2\)

Step2: Oxidation state of C in \(CO_3^{2-}\)

Let the oxidation state of C be \(y\).
The oxidation state of O is \(-2\).
Using the formula: \(y+3\times(-2)=-2\) (since the overall charge of \(CO_3^{2-}\) is \(-2\))
\(y-6=-2\)
\(y=-2 + 6=+4\)

Step3: Oxidation state of Cr in \(CrO_4^{2-}\)

Let the oxidation state of Cr be \(z\).
The oxidation state of O is \(-2\).
Using the formula: \(z+4\times(-2)=-2\) (since the overall charge of \(CrO_4^{2-}\) is \(-2\))
\(z-8=-2\)
\(z=-2 + 8=+6\)

Step4: Oxidation state of C in \(HCO_3^{-}\)

Let the oxidation state of C be \(w\).
The oxidation state of H is \(+1\) and of O is \(-2\).
Using the formula: \(+1+w+3\times(-2)=-1\) (since the overall charge of \(HCO_3^{-}\) is \(-1\))
\(1+w - 6=-1\)
\(w=-1+6 - 1=+4\)

Answer:

species	oxidation state of highlighted atom
\(CO_3^{2-}(aq)\)	\(+4\)
\(CrO_4^{2-}(aq)\)	\(+6\)
\(HCO_3^{-}(aq)\)	\(+4\)