Question

the two balanced half-reactions are given below.
\\(3\mathrm{e}^- + 2\mathrm{h}_2\mathrm{o} + \mathrm{mno}_4^- \
ightarrow \mathrm{mno}_2 + 4\mathrm{oh}^-\\)
\\(6\mathrm{oh}^- + \mathrm{br}^- \
ightarrow \mathrm{bro}_3^- + 3\mathrm{h}_2\mathrm{o} + 6\mathrm{e}^-\\)
now that both half-reactions have been balanced, a matching number of electrons is needed to combine the two.
what value does the top half-reaction need to be multiplied by for both reactions to have a total of 6 electrons?
\\(?(3\mathrm{e}^- + 2\mathrm{h}_2\mathrm{o} + \mathrm{mno}_4^- \
ightarrow \mathrm{mno}_2 + 4\mathrm{oh}^-)\\)

Explanation:

Step1: Identify electrons in top reaction

Top half - reaction has \(3e^-\).

Step2: Calculate multiplier for 6 electrons

Let the multiplier be \(x\). We want \(3x = 6\). Solving for \(x\), we divide both sides by 3: \(x=\frac{6}{3}=2\).

Answer:

2