Question

the two half-reactions are given below.
\\( 3\text{e}^- + 4\text{h}_2\text{o} + \text{tco}_4^- \
ightarrow \text{tc}^{+4} + 8\text{oh}^- \\)
\\( \text{sn}^{+2} \
ightarrow \text{sn}^{+4} + 2\text{e}^- \\)
what is the coefficient for \\( \text{h}_2\text{o} \\) in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in half - reactions

The first half - reaction (reduction) involves 3 electrons: \(3e^-+4H_2O + TcO_4^-
ightarrow Tc^{+4}+8OH^-\)
The second half - reaction (oxidation) involves 2 electrons: \(Sn^{+2}
ightarrow Sn^{+4}+2e^-\)

To balance the number of electrons, we find the least common multiple of 3 and 2, which is 6.

Multiply the first half - reaction by 2: \(2\times(3e^- + 4H_2O+TcO_4^-
ightarrow Tc^{+4}+8OH^-)\) gives \(6e^-+8H_2O + 2TcO_4^-
ightarrow 2Tc^{+4}+16OH^-\)

Multiply the second half - reaction by 3: \(3\times(Sn^{+2}
ightarrow Sn^{+4}+2e^-)\) gives \(3Sn^{+2}
ightarrow 3Sn^{+4}+6e^-\)

Step2: Combine the two half - reactions

Now, we add the two modified half - reactions together. The electrons (\(6e^-\)) will cancel out.

\(6e^-+8H_2O + 2TcO_4^-+3Sn^{+2}
ightarrow 2Tc^{+4}+16OH^-+3Sn^{+4}+6e^-\)

After canceling the electrons, the equation is \(8H_2O + 2TcO_4^-+3Sn^{+2}
ightarrow 2Tc^{+4}+16OH^-+3Sn^{+4}\)

Answer:

The coefficient of \(H_2O\) in the balanced net ionic equation is 8.