Question

the redox reaction given below occurs in basic solution.\\(\ce{mno^-_{4} + cn^- \
ightarrow cno^- + mno_{2}}\\)\
balance the half-reaction:\\(\ce{mno^-_{4} \
ightarrow mno_{2}}\\)\
how many electrons are transferred?\\(\ce{?e^-}\\)

Explanation:

Step1: Determine oxidation states of Mn

In \( \text{MnO}_4^- \), let oxidation state of Mn be \( x \). Oxygen is -2, so \( x + 4(-2) = -1 \). Solving, \( x - 8 = -1 \) → \( x = +7 \). In \( \text{MnO}_2 \), let oxidation state of Mn be \( y \). Then \( y + 2(-2) = 0 \) → \( y - 4 = 0 \) → \( y = +4 \).

Step2: Calculate electron transfer

Mn goes from +7 to +4, so it gains electrons. The change in oxidation state is \( 7 - 4 = 3 \). So each Mn atom gains 3 electrons. Since the half - reaction is \( \text{MnO}_4^-
ightarrow \text{MnO}_2 \), and we're balancing for Mn (which is 1:1 here), the number of electrons transferred per Mn atom is 3.

Answer:

3