Answer:
The net ionic equation is as follows:
HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)
Explanation:
The reaction between Hydrocyanic acid, HCN, and sodium hydroxide is a neutralization reaction between a weak acid and a strong base.
Hydrocyanic acid being a weak acid ionizes only slightly, while sodium hydroxide being a strong base ionizes completely. The equation for the reaction is given below:
A. HCN(aq) + NaOH-(aq) ----> NaCN(aq) + H2O(l)
Since Hydrocyanic acid is written in the aqueous form as it ionizes only slightly and the ionic equation is given below:
HCN(aq) + Na+(aq)+OH-(aq) ----> Na+(aq)+CN-(aq) + H2O(l)
Na+ being a spectator ion is removed from the net ionic equation given below:
HCN(aq) + OH-(aq) ----> H20(l) + CN-(aq)
<span>Average oxidation state = VO1.19
Oxygen is-2. Then 1.19 (-2) = -2.38
Average oxidation state of V is +2.38
Consider 100 formula units of VO1.19
There would be 119 Oxide ions = Each oxide is -2. Total charge = -2(119) = -238
The total charge of all the vanadium ions would be +238.
Let x = number of of V+2
Then 100 – x = number of V+3
X(+2) + 100-x(+3) = +238
2x + 300 – 3x = 238
-x = 238-300 = -62
x = 62
Thus 62/100 are V+2
62/100 * 100 = 62%
</span>62 % is the percentage of the vanadium atoms are in the lower oxidation state. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
