Question

Depending upon the conditions, potassium permanganate may form Mn(s), Mn²⁺, or Mn³⁺. Write a balanced half-reaction chemical equation for each reaction. Determine the number of electrons transferred and the equivalent mass of potassium permanganate for each condition. Also determine the molarity of a 0.200 M solution of KMnO₄ assuming the final oxidation state of manganese if +2.

Answer 1

1. MnO₄^- (aq) + 8H⁺ (aq) + 7e^- = Mn (s) + 4H₂O (l)

2. MnO₄^- (aq) + 8H⁺ (aq) + 4e^- = Mn³⁺ (aq) + 4H₂O (l)

3. MnO₄^- (aq) + 8H⁺ (aq) + 5e^- = Mn²⁺ (aq) + 4H₂O (l)

equivalent mass is:

M_eq = M/n_el

M - molar mass

n_el - number of electrons

1. M_eq = M/n_el = 158.034 g/mol / 7 = 22.58 g / mol-eq

2. M_eq = M/n_el = 158.034 g/mol / 4 = 39.51 g / mol-eq

3. M_eq = M/n_el = 158.034 g/mol / 5 = 31.61 g / mol-eq

c(mol/L) = n(mol)/V(L) = m(g)/(M(g/mol)*V(L)

c(mol-eq/L) = n(mol-eq)/V(L) = m(g)/(M(g/mol-eq)*V(L)

m(g)/V(L) = const

c(mol/L)*M(g/mol) = c(mol-eq/L)*M(g/mol-eq)

c(mol-eq/L) = c(mol/L)*M(g/mol)/M(g/mol-eq)

c(mol-eq/L) = 0.2 mol/L * 158.034 g/mol / 31.61 g / mol-eq = 1.0 mol-eq/L = 1 H