Question

the redox reaction given below occurs in basic solution.
mno₄⁻ + cn⁻ → cno⁻ + mno₂
balance the half - reaction:
cn⁻ → cno⁻
how many electrons are transferred?
?e⁻

Explanation:

Step1: Determine oxidation states

In \( \text{CN}^- \), let the oxidation state of \( \text{C} \) be \( x \), \( x + (-3) = -1 \), so \( x = +2 \). In \( \text{CNO}^- \), let the oxidation state of \( \text{C} \) be \( y \), \( y + (-3) + (-2) = -1 \), so \( y = +4 \).

Step2: Balance atoms other than O and H

\( \text{CN}^-
ightarrow \text{CNO}^- \) (C and N are balanced).

Step3: Balance O by adding \( \text{H}_2\text{O} \)

Add 1 \( \text{H}_2\text{O} \) to the left: \( \text{H}_2\text{O} + \text{CN}^-
ightarrow \text{CNO}^- \).

Step4: Balance H by adding \( \text{H}^+ \)

Add 2 \( \text{H}^+ \) to the right: \( \text{H}_2\text{O} + \text{CN}^-
ightarrow \text{CNO}^- + 2\text{H}^+ \).

Step5: Adjust for basic solution (add \( \text{OH}^- \) to both sides)

Add 2 \( \text{OH}^- \) to both sides: \( 2\text{OH}^- + \text{H}_2\text{O} + \text{CN}^-
ightarrow \text{CNO}^- + 2\text{H}^+ + 2\text{OH}^- \). The \( 2\text{H}^+ \) and \( 2\text{OH}^- \) form \( 2\text{H}_2\text{O} \), so: \( 2\text{OH}^- + \text{CN}^-
ightarrow \text{CNO}^- + \text{H}_2\text{O} \).

Step6: Balance charge by adding electrons

Left side charge: \( -2 + (-1) = -3 \). Right side charge: \( -1 + 0 = -1 \). To balance charge, add 2 \( e^- \) to the right: \( 2\text{OH}^- + \text{CN}^-
ightarrow \text{CNO}^- + \text{H}_2\text{O} + 2e^- \). The change in oxidation state of C is \( +4 - +2 = +2 \), so 2 electrons are transferred.

Answer:

2