Question

use the references to access important values if needed for this question.
when the following skeletal equation is balanced under acidic conditions, what are the coefficients of the species shown?
\\( \square \mathrm{sb}_{2} \mathrm{o}_{5}+\square \mathrm{mn} \
ightarrow \square \mathrm{sbo}^{+}+\square \mathrm{mn}^{2+} \\)
water appears in the balanced equation as a (reactant, product, neither) with a coefficient of (enter 0 for neither.)
which species is the reducing agent?

Explanation:

Step1: Assign oxidation numbers

• For \(Sb_2O_5\): \(Sb\) has an oxidation number of \(+5\) (since \(O\) is \(- 2\), and \(2x+5\times(-2)=0\), \(x = + 5\)). In \(SbO^+\), \(Sb\) has an oxidation number of \(+3\) (since \(O\) is \(-2\), \(x+(-2)=+1\), \(x = + 3\)).
• For \(Mn\): oxidation number is \(0\). In \(Mn^{2+}\), oxidation number is \(+2\).

Step2: Write half - reactions

Oxidation half - reaction:
\(Mn
ightarrow Mn^{2+}\)
Balance charge: \(Mn
ightarrow Mn^{2+}+2e^-\)

Reduction half - reaction:
\(Sb_2O_5
ightarrow SbO^+\)
Balance \(Sb\): \(Sb_2O_5
ightarrow 2SbO^+\)
Balance \(O\) (in acidic medium, add \(H_2O\)): \(Sb_2O_5
ightarrow 2SbO^++H_2O\)
Balance \(H\) (add \(H^+\)): \(Sb_2O_5 + 2H^+
ightarrow 2SbO^++H_2O\)
Balance charge: \(Sb_2O_5 + 2H^++4e^-
ightarrow 2SbO^++H_2O\)

Step3: Equalize electrons

Multiply oxidation half - reaction by \(2\): \(2Mn
ightarrow 2Mn^{2+}+4e^-\)
Add to reduction half - reaction: \(Sb_2O_5 + 2H^++2Mn
ightarrow 2SbO^++H_2O+2Mn^{2+}\)

Answer:

The coefficients: \(Sb_2O_5 = 1\), \(Mn = 2\), \(SbO^+=2\), \(Mn^{2+}=2\)
Water appears as a product with a coefficient of \(1\)
The reducing agent is \(Mn\)