1. First reaction: $2Fe^{2+} + H_2O_2 \to xA + yB$ in basic medium

In basic medium, $Fe^{2+}$ is oxidized to $Fe^{3+}$ and $H_2O_2$ is reduced to $OH^-$.

Let us write the half-reactions.

• Oxidation: $Fe^{2+} \to Fe^{3+} + e^-$ So for 2 iron ions, $2Fe^{2+} \to 2Fe^{3+} + 2e^-$

• Reduction of hydrogen peroxide in basic medium: $H_2O_2 + 2e^- \to 2OH^-$

Adding them, $2Fe^{2+} + H_2O_2 \to 2Fe^{3+} + 2OH^-$

Hence, $A = Fe^{3+},\quad B = OH^-$ So, $x=2,\quad y=2$

2. Second reaction: $2MnO_4^- + 6H^+ + 5H_2O_2 \to x'C + y'D + z'E$ in acidic medium

In acidic medium, permanganate is reduced to $Mn^{2+}$ and hydrogen peroxide is oxidized to oxygen.

Half-reactions:

• Reduction: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$ Multiplying by 2: $2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$

• Oxidation: $H_2O_2 \to O_2 + 2H^+ + 2e^-$ Multiplying by 5: $5H_2O_2 \to 5O_2 + 10H^+ + 10e^-$

Adding: $2MnO_4^- + 16H^+ + 5H_2O_2 \to 2Mn^{2+} + 8H_2O + 5O_2 + 10H^+$

Cancel $10H^+$ from both sides: $2MnO_4^- + 6H^+ + 5H_2O_2 \to 2Mn^{2+} + 8H_2O + 5O_2$

Thus, $C = Mn^{2+},\quad D = H_2O,\quad E = O_2$ So, $x' = 2,\quad y' = 8,\quad z' = 5$

3. Required sum

$x+y+x'+y'+z' = 2+2+2+8+5 = 19$

4. Comparison with stored answer

Stored correct answer = 19.

Our derived answer also is 19, so it agrees.