Question

the two half-reactions are given below.
$6\mathrm{e}^- + 7\mathrm{h}_2\mathrm{o} + \mathrm{cr}_2\mathrm{o}_7^{-2} \
ightarrow 2\mathrm{cr}^{+3} + 14\mathrm{oh}^-$
$12\mathrm{oh}^- + \mathrm{i}_2 \
ightarrow 2\mathrm{io}_3^- + 6\mathrm{h}_2\mathrm{o} + 10\mathrm{e}^-$
what is the coefficient for $\mathrm{cr}^{+3}$ in the combined balanced net ionic equation?

Explanation:

Step1: Find the least common multiple of electrons transferred

The first half - reaction (reduction) involves the transfer of 6 electrons, and the second half - reaction (oxidation) involves the transfer of 10 electrons. The least common multiple of 6 and 10 is 30.

Step2: Multiply the half - reactions to balance electrons

Multiply the first half - reaction (reduction: \(6e^- + 7H_2O+Cr_2O_7^{2 - }
ightarrow2Cr^{3 + }+14OH^-\)) by 5 to get \(30e^-+35H_2O + 5Cr_2O_7^{2-}
ightarrow10Cr^{3+}+70OH^-\)
Multiply the second half - reaction (oxidation: \(12OH^-+I_2
ightarrow2IO_3^-+6H_2O + 10e^-\)) by 3 to get \(36OH^-+3I_2
ightarrow6IO_3^-+18H_2O+30e^-\)

Step3: Add the two modified half - reactions

Add the two equations:
\((30e^-+35H_2O + 5Cr_2O_7^{2-})+(36OH^-+3I_2)=(10Cr^{3+}+70OH^-)+(6IO_3^-+18H_2O+30e^-)\)
Cancel out the electrons (\(30e^-\) on both sides). Then simplify the water and hydroxide ions:
For water: \(35H_2O-18H_2O = 17H_2O\)
For hydroxide ions: \(36OH^--70OH^-=- 34OH^-\)
The equation becomes: \(5Cr_2O_7^{2-}+3I_2 + 17H_2O=10Cr^{3+}+6IO_3^-+34OH^-\)
From the balanced equation, we can see that the coefficient of \(Cr^{3+}\) is 10.

Answer:

10