(a) The balanced overall equation for the reaction is:
$5H_{2}O_{2}(aq)+2MnO_{4}^{-}(aq)+16H^{+}(aq)\rightarrow 5O_{2}(g)+2Mn^{2+}(aq)+8H_{2}O(l)$
To calculate E cell, we can use the formula:
$E_{cell} = E_{cathode} - E_{anode}$
The reduction half-reaction is $MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-} \rightarrow Mn^{2+}(aq) + 4H_{2}O(l)$, with $E_{red}^{\circ} = 1.51V$.
The oxidation half-reaction is $5H_{2}O_{2}(aq) \rightarrow 5O_{2}(g) + 10H^{+}(aq) + 10e^{-}$, with $E_{red}^{\circ} = +0.68V$.
Therefore,
$E_{cell} = 1.51V - 0.68V = 0.83V$
(b) The balanced chemical equation shows that 2 moles of MnO$_4^{-}$ react with 5 moles of $H_2O_2$.
Given that the volume of $KMnO_4$ used is 21.24 ml and the concentration is 0.01 M,
number of moles of $KMnO_4 = 0.01 \times 21.24 \times 10^{-3} = 2.124 \times 10^{-4}$ moles
Since 2 moles of $KMnO_4$ react with 5 moles of $H_2O_2$,
number of moles of $H_2O_2 = 2.124 \times 10^{-4} \times \frac{5}{2} = 5.31 \times 10^{-4}$ moles
Volume of the undiluted hydrogen peroxide used = 20 ml
Concentration of original $H_2O_2 = \frac{5.31 \times 10^{-4}}{20 \times 10^{-3}} = 0.02655 M$
Q1. A 25 mL sample of commercial hydrogen peroxide was diluted to 250 mL.Then 20 mL of this diluted solution was acidified and titrated with a 0.01 M potassium permanganate solution.An average volume of 21.24 ml of potassium permanganate solution was required according to the following half equations: (8 marks)
$O_{2}(g)+2H^{+}(aq)+2e\rightarrow H_{2}O_{2}(aq)$ $MnO_{4}^{-}(aq)+8H^{+}(aq)+5e\rightarrow Mn^{2+}(aq)+4H_{2}O(l)$
$E_{red}^{\circ }=+0.68V$ $E_{red}^{\circ }=+1.51V$
(a) Write a balanced overall equation for the reaction and calculate E cell.(4 marks) (b) Calculate the concentration of the original hydrogen peroxide. (4 marks)
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